Is there any way not to use l'Hôpital's rule when computing $lim_{xto8}frac{sqrt[3]x-2}{x-8}$? [on hold]Is...

What does a comma signify in inorganic chemistry?

Why should P.I be willing to write strong LOR even if that means losing a undergraduate from his/her lab?

Why don't modern jet engines use forced exhaust mixing?

Why can't I see 1861 / 1871 census entries on Freecen website when I can see them on Ancestry website?

What is the purpose/function of this power inductor in parallel?

The anatomy of an organic infrared generator

What's the relationship betweeen MS-DOS and XENIX?

Does the Temple of the Gods spell nullify critical hits?

Have made several mistakes during the course of my PhD. Can't help but feel resentment. Can I get some advice about how to move forward?

Vegetarian dishes on Russian trains (European part)

Photoshop older default brushes

Reducing contention in thread-safe LruCache

Adding things to bunches of things vs multiplication

How to train a replacement without them knowing?

May the tower use the runway while an emergency aircraft is inbound?

If it isn't [someone's name]!

What modifiers are added to the attack and damage rolls of this unique longbow from Waterdeep: Dragon Heist?

global variant of csname…endcsname

What's the point of writing that I know will never be used or read?

Meaning and structure of headline "Hair it is: A List of ..."

Number of matrices with bounded products of rows and columns

What is the best way to use errors in mathematica M12?

What allows us to use imaginary numbers?

How to render "have ideas above his station" into German



Is there any way not to use l'Hôpital's rule when computing $lim_{xto8}frac{sqrt[3]x-2}{x-8}$? [on hold]


Is this use of L'Hôpital's rule incorrect?Is the use of L'Hôpital's rule in this limit wrong?When to Use L'Hôpital's RuleCalculate $lim_{x to 1^{-}} frac{arccos{x}}{sqrt{1-x}}$ without using L'Hôpital's rule.$lim_{xrightarrow 6} frac{sqrt{x+3}-3}{x-6}$ without L'Hôpital's ruleSolve $lim_{x rightarrow 1}frac{sqrt x -1}{2-sqrt{x+3}}$ without using L'Hôpital's rule.Use L'Hôpital's rule to solve $lim_{xto 0^{+}}sin(x)ln(x)$How many times do you have to use L'Hôpital's rule?How can I calculate $lim_{x to 0}frac {cos x- sqrt {cos 2x}×sqrt[3] {cos 3x}}{x^2}$ without L'Hôpital's rule?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}







1












$begingroup$


$$lim_{xto8}frac{sqrt[3]x-2}{x-8}$$
Is there any way I will not use l'Hôpital's rule here?






calculus limits limits-without-lhopital







share|cite|improve this question









New contributor



Jhon Yrap Masuli Gahiton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




put on hold as off-topic by RRL, TheSimpliFire, Paul Frost, Ak19, Daniele Tampieri yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, TheSimpliFire, Paul Frost, Ak19, Daniele Tampieri

If this question can be reworded to fit the rules in the help center, please edit the question.













  • 1




    $begingroup$
    Just to confirm: is your question 'is there any way to not use l'Hopital's rule?' Maybe you can show what you tried using l'Hopital's rule?
    $endgroup$
    – Toby Mak
    2 days ago












  • $begingroup$
    How do you suppose we define $f'(8)$ where $f(x)=sqrt[3]{x}$?
    $endgroup$
    – Peter Foreman
    2 days ago


















1












$begingroup$


$$lim_{xto8}frac{sqrt[3]x-2}{x-8}$$
Is there any way I will not use l'Hôpital's rule here?






calculus limits limits-without-lhopital







share|cite|improve this question









New contributor



Jhon Yrap Masuli Gahiton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




put on hold as off-topic by RRL, TheSimpliFire, Paul Frost, Ak19, Daniele Tampieri yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, TheSimpliFire, Paul Frost, Ak19, Daniele Tampieri

If this question can be reworded to fit the rules in the help center, please edit the question.













  • 1




    $begingroup$
    Just to confirm: is your question 'is there any way to not use l'Hopital's rule?' Maybe you can show what you tried using l'Hopital's rule?
    $endgroup$
    – Toby Mak
    2 days ago












  • $begingroup$
    How do you suppose we define $f'(8)$ where $f(x)=sqrt[3]{x}$?
    $endgroup$
    – Peter Foreman
    2 days ago














1












1








1





$begingroup$


$$lim_{xto8}frac{sqrt[3]x-2}{x-8}$$
Is there any way I will not use l'Hôpital's rule here?






calculus limits limits-without-lhopital







share|cite|improve this question









New contributor



Jhon Yrap Masuli Gahiton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




$$lim_{xto8}frac{sqrt[3]x-2}{x-8}$$
Is there any way I will not use l'Hôpital's rule here?






calculus limits limits-without-lhopital




calculus limits limits-without-lhopital






share|cite|improve this question









New contributor



Jhon Yrap Masuli Gahiton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










share|cite|improve this question









New contributor



Jhon Yrap Masuli Gahiton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Asaf Karagila♦

315k35 gold badges454 silver badges788 bronze badges




315k35 gold badges454 silver badges788 bronze badges






New contributor



Jhon Yrap Masuli Gahiton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








asked 2 days ago









Jhon Yrap Masuli GahitonJhon Yrap Masuli Gahiton

193 bronze badges




193 bronze badges




New contributor



Jhon Yrap Masuli Gahiton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




New contributor




Jhon Yrap Masuli Gahiton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







put on hold as off-topic by RRL, TheSimpliFire, Paul Frost, Ak19, Daniele Tampieri yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, TheSimpliFire, Paul Frost, Ak19, Daniele Tampieri

If this question can be reworded to fit the rules in the help center, please edit the question.









put on hold as off-topic by RRL, TheSimpliFire, Paul Frost, Ak19, Daniele Tampieri yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, TheSimpliFire, Paul Frost, Ak19, Daniele Tampieri

If this question can be reworded to fit the rules in the help center, please edit the question.







put on hold as off-topic by RRL, TheSimpliFire, Paul Frost, Ak19, Daniele Tampieri yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, TheSimpliFire, Paul Frost, Ak19, Daniele Tampieri

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    Just to confirm: is your question 'is there any way to not use l'Hopital's rule?' Maybe you can show what you tried using l'Hopital's rule?
    $endgroup$
    – Toby Mak
    2 days ago












  • $begingroup$
    How do you suppose we define $f'(8)$ where $f(x)=sqrt[3]{x}$?
    $endgroup$
    – Peter Foreman
    2 days ago














  • 1




    $begingroup$
    Just to confirm: is your question 'is there any way to not use l'Hopital's rule?' Maybe you can show what you tried using l'Hopital's rule?
    $endgroup$
    – Toby Mak
    2 days ago












  • $begingroup$
    How do you suppose we define $f'(8)$ where $f(x)=sqrt[3]{x}$?
    $endgroup$
    – Peter Foreman
    2 days ago








1




1




$begingroup$
Just to confirm: is your question 'is there any way to not use l'Hopital's rule?' Maybe you can show what you tried using l'Hopital's rule?
$endgroup$
– Toby Mak
2 days ago






$begingroup$
Just to confirm: is your question 'is there any way to not use l'Hopital's rule?' Maybe you can show what you tried using l'Hopital's rule?
$endgroup$
– Toby Mak
2 days ago














$begingroup$
How do you suppose we define $f'(8)$ where $f(x)=sqrt[3]{x}$?
$endgroup$
– Peter Foreman
2 days ago




$begingroup$
How do you suppose we define $f'(8)$ where $f(x)=sqrt[3]{x}$?
$endgroup$
– Peter Foreman
2 days ago










6 Answers
6






active

oldest

votes


















7












$begingroup$

The denominator can be factored as
$$(x^{1/3}-2)(x^{2/3}+2x^{1/3}+4)$$
so
$$lim_{xto8}frac{x^{1/3}-2}{x-8}=lim_{xto8}frac1{x^{2/3}+2x^{1/3}+4}=frac1{12}$$






share|cite|improve this answer









$endgroup$















  • $begingroup$
    this one but I cant understand exponents in fraction can you make this simplier for me to understand sorry for ignorance
    $endgroup$
    – Jhon Yrap Masuli Gahiton
    2 days ago










  • $begingroup$
    @JhonYrapMasuliGahiton Basically, $x^{1/3}=sqrt[3]x$ and $x^{2/3}=sqrt[3]{x^2}$.
    $endgroup$
    – Parcly Taxel
    2 days ago






  • 1




    $begingroup$
    I get it now problem solved!
    $endgroup$
    – Jhon Yrap Masuli Gahiton
    2 days ago



















6












$begingroup$

Hint.



$$
frac{sqrt[3]{x}-2}{x-8} = frac{sqrt[3]{x}-2}{(sqrt[3]{x})^3-2^3}
$$






share|cite|improve this answer









$endgroup$















  • $begingroup$
    can you show me the steps hehe badly needed help
    $endgroup$
    – Jhon Yrap Masuli Gahiton
    2 days ago










  • $begingroup$
    i know the answer is 1/12 but the one I found uses theorem we havent tackled it yet in our class
    $endgroup$
    – Jhon Yrap Masuli Gahiton
    2 days ago






  • 3




    $begingroup$
    $frac{a-b}{a^3-b^3} = frac{a-b}{(a-b)(a^2+ab+b^2)} = frac{1}{a^2+ab+b^2}$.
    $endgroup$
    – Toby Mak
    2 days ago





















6












$begingroup$

Hint$$a^3-b^3=(a-b)(a^2+ab+b^2)$$






share|cite|improve this answer









$endgroup$























    5












    $begingroup$

    From Peter Foreman's comment



    By the definition of a derivative:



    $$lim_{a to 8} frac{a^{1/3} - 8^{1/3}}{a-8}$$
    $$= frac{mathrm d}{mathrm da} left( a^{1/3} right) biggl|_{a=8}$$






    share|cite|improve this answer











    $endgroup$























      4












      $begingroup$

      Recall that there are two equivalent definitions of the derivative of a function in calculus. We're interested in this one:



      $$f'(a)=lim_{xto a}frac{f(x)-f(a)}{x-a}.$$



      Your limit can be easily rewritten to fit that form. The only other thing that you would need here is the derivative of $sqrt[3]{x}$. It's quite easy to find using the so-called power rule. It's equal to $frac{1}{3sqrt[3]{x^2}}$. So, your problem now becomes the problem of evaluating the derivative of the function $sqrt[3]{x}$ at $x=8$:



      $$
      lim_{xto 8}frac{sqrt[3]{x}-2}{x-8}=
      lim_{xto 8}frac{sqrt[3]{x}-sqrt[3]{8}}{x-8}=
      frac{d}{dx}left(sqrt[3]{x}right)bigg|_{x=8}=\
      frac{1}{3sqrt[3]{x^2}}bigg|_{x=8}=
      frac{1}{3sqrt[3]{8^2}}=
      frac{1}{3cdot 4}=
      frac{1}{12}.
      $$






      share|cite|improve this answer











      $endgroup$











      • 1




        $begingroup$
        this is great but we are not yet on derivatives please without using derivatives I cant understand yet
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago



















      0












      $begingroup$

      There is a standard binomial limit which you may use
      Limit x tend to a (x^n-a^n)/(x-a) = na^(n-1)
      This is proved by using the maclaurin series expansions as far I as can recall...the answer by this method for your problem would be (1/12)






      share|cite|improve this answer








      New contributor



      Schwarz Kugelblitz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      $endgroup$















      • $begingroup$
        This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
        $endgroup$
        – Klangen
        2 days ago










      • $begingroup$
        I fail to see how this doesn't answer the question.... this is a standard formula taught in India for competitive entrance exams like the JEE (joint entrance exam) with a decent amount of proofs behind it and a good number of authors publishing it in their books...the person who asked the question wanted an alternate approach other than l'hopital and I gave a valid mathematically sound one
        $endgroup$
        – Schwarz Kugelblitz
        2 days ago






      • 2




        $begingroup$
        Two things that would make this look more like an answer: give a reason for the formula that is better than "it's in a bunch of test prep books" (a proof of the formula would be such a reason), and format the formula in standard notation so that people can more easily recognize it: see math.meta.stackexchange.com/questions/5020/…
        $endgroup$
        – David K
        2 days ago












      • $begingroup$
        What you would have to do is to show that $frac{x^n-a^n}{x-a} = x^n+x^{n-1}a + cdots + xa^{n-1} + a^n$, and then substitute $x=a$ to get to your answer. Bear in mind that this question might also be useful to others, who might not take JEE and read the books you are suggesting.
        $endgroup$
        – Toby Mak
        2 days ago






















      6 Answers
      6






      active

      oldest

      votes








      6 Answers
      6






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$

      The denominator can be factored as
      $$(x^{1/3}-2)(x^{2/3}+2x^{1/3}+4)$$
      so
      $$lim_{xto8}frac{x^{1/3}-2}{x-8}=lim_{xto8}frac1{x^{2/3}+2x^{1/3}+4}=frac1{12}$$






      share|cite|improve this answer









      $endgroup$















      • $begingroup$
        this one but I cant understand exponents in fraction can you make this simplier for me to understand sorry for ignorance
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago










      • $begingroup$
        @JhonYrapMasuliGahiton Basically, $x^{1/3}=sqrt[3]x$ and $x^{2/3}=sqrt[3]{x^2}$.
        $endgroup$
        – Parcly Taxel
        2 days ago






      • 1




        $begingroup$
        I get it now problem solved!
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago
















      7












      $begingroup$

      The denominator can be factored as
      $$(x^{1/3}-2)(x^{2/3}+2x^{1/3}+4)$$
      so
      $$lim_{xto8}frac{x^{1/3}-2}{x-8}=lim_{xto8}frac1{x^{2/3}+2x^{1/3}+4}=frac1{12}$$






      share|cite|improve this answer









      $endgroup$















      • $begingroup$
        this one but I cant understand exponents in fraction can you make this simplier for me to understand sorry for ignorance
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago










      • $begingroup$
        @JhonYrapMasuliGahiton Basically, $x^{1/3}=sqrt[3]x$ and $x^{2/3}=sqrt[3]{x^2}$.
        $endgroup$
        – Parcly Taxel
        2 days ago






      • 1




        $begingroup$
        I get it now problem solved!
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago














      7












      7








      7





      $begingroup$

      The denominator can be factored as
      $$(x^{1/3}-2)(x^{2/3}+2x^{1/3}+4)$$
      so
      $$lim_{xto8}frac{x^{1/3}-2}{x-8}=lim_{xto8}frac1{x^{2/3}+2x^{1/3}+4}=frac1{12}$$






      share|cite|improve this answer









      $endgroup$



      The denominator can be factored as
      $$(x^{1/3}-2)(x^{2/3}+2x^{1/3}+4)$$
      so
      $$lim_{xto8}frac{x^{1/3}-2}{x-8}=lim_{xto8}frac1{x^{2/3}+2x^{1/3}+4}=frac1{12}$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 2 days ago









      Parcly TaxelParcly Taxel

      51.5k13 gold badges80 silver badges120 bronze badges




      51.5k13 gold badges80 silver badges120 bronze badges















      • $begingroup$
        this one but I cant understand exponents in fraction can you make this simplier for me to understand sorry for ignorance
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago










      • $begingroup$
        @JhonYrapMasuliGahiton Basically, $x^{1/3}=sqrt[3]x$ and $x^{2/3}=sqrt[3]{x^2}$.
        $endgroup$
        – Parcly Taxel
        2 days ago






      • 1




        $begingroup$
        I get it now problem solved!
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago


















      • $begingroup$
        this one but I cant understand exponents in fraction can you make this simplier for me to understand sorry for ignorance
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago










      • $begingroup$
        @JhonYrapMasuliGahiton Basically, $x^{1/3}=sqrt[3]x$ and $x^{2/3}=sqrt[3]{x^2}$.
        $endgroup$
        – Parcly Taxel
        2 days ago






      • 1




        $begingroup$
        I get it now problem solved!
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago
















      $begingroup$
      this one but I cant understand exponents in fraction can you make this simplier for me to understand sorry for ignorance
      $endgroup$
      – Jhon Yrap Masuli Gahiton
      2 days ago




      $begingroup$
      this one but I cant understand exponents in fraction can you make this simplier for me to understand sorry for ignorance
      $endgroup$
      – Jhon Yrap Masuli Gahiton
      2 days ago












      $begingroup$
      @JhonYrapMasuliGahiton Basically, $x^{1/3}=sqrt[3]x$ and $x^{2/3}=sqrt[3]{x^2}$.
      $endgroup$
      – Parcly Taxel
      2 days ago




      $begingroup$
      @JhonYrapMasuliGahiton Basically, $x^{1/3}=sqrt[3]x$ and $x^{2/3}=sqrt[3]{x^2}$.
      $endgroup$
      – Parcly Taxel
      2 days ago




      1




      1




      $begingroup$
      I get it now problem solved!
      $endgroup$
      – Jhon Yrap Masuli Gahiton
      2 days ago




      $begingroup$
      I get it now problem solved!
      $endgroup$
      – Jhon Yrap Masuli Gahiton
      2 days ago













      6












      $begingroup$

      Hint.



      $$
      frac{sqrt[3]{x}-2}{x-8} = frac{sqrt[3]{x}-2}{(sqrt[3]{x})^3-2^3}
      $$






      share|cite|improve this answer









      $endgroup$















      • $begingroup$
        can you show me the steps hehe badly needed help
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago










      • $begingroup$
        i know the answer is 1/12 but the one I found uses theorem we havent tackled it yet in our class
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago






      • 3




        $begingroup$
        $frac{a-b}{a^3-b^3} = frac{a-b}{(a-b)(a^2+ab+b^2)} = frac{1}{a^2+ab+b^2}$.
        $endgroup$
        – Toby Mak
        2 days ago


















      6












      $begingroup$

      Hint.



      $$
      frac{sqrt[3]{x}-2}{x-8} = frac{sqrt[3]{x}-2}{(sqrt[3]{x})^3-2^3}
      $$






      share|cite|improve this answer









      $endgroup$















      • $begingroup$
        can you show me the steps hehe badly needed help
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago










      • $begingroup$
        i know the answer is 1/12 but the one I found uses theorem we havent tackled it yet in our class
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago






      • 3




        $begingroup$
        $frac{a-b}{a^3-b^3} = frac{a-b}{(a-b)(a^2+ab+b^2)} = frac{1}{a^2+ab+b^2}$.
        $endgroup$
        – Toby Mak
        2 days ago
















      6












      6








      6





      $begingroup$

      Hint.



      $$
      frac{sqrt[3]{x}-2}{x-8} = frac{sqrt[3]{x}-2}{(sqrt[3]{x})^3-2^3}
      $$






      share|cite|improve this answer









      $endgroup$



      Hint.



      $$
      frac{sqrt[3]{x}-2}{x-8} = frac{sqrt[3]{x}-2}{(sqrt[3]{x})^3-2^3}
      $$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 2 days ago









      CesareoCesareo

      12.4k3 gold badges6 silver badges21 bronze badges




      12.4k3 gold badges6 silver badges21 bronze badges















      • $begingroup$
        can you show me the steps hehe badly needed help
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago










      • $begingroup$
        i know the answer is 1/12 but the one I found uses theorem we havent tackled it yet in our class
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago






      • 3




        $begingroup$
        $frac{a-b}{a^3-b^3} = frac{a-b}{(a-b)(a^2+ab+b^2)} = frac{1}{a^2+ab+b^2}$.
        $endgroup$
        – Toby Mak
        2 days ago




















      • $begingroup$
        can you show me the steps hehe badly needed help
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago










      • $begingroup$
        i know the answer is 1/12 but the one I found uses theorem we havent tackled it yet in our class
        $endgroup$
        – Jhon Yrap Masuli Gahiton
        2 days ago






      • 3




        $begingroup$
        $frac{a-b}{a^3-b^3} = frac{a-b}{(a-b)(a^2+ab+b^2)} = frac{1}{a^2+ab+b^2}$.
        $endgroup$
        – Toby Mak
        2 days ago


















      $begingroup$
      can you show me the steps hehe badly needed help
      $endgroup$
      – Jhon Yrap Masuli Gahiton
      2 days ago




      $begingroup$
      can you show me the steps hehe badly needed help
      $endgroup$
      – Jhon Yrap Masuli Gahiton
      2 days ago












      $begingroup$
      i know the answer is 1/12 but the one I found uses theorem we havent tackled it yet in our class
      $endgroup$
      – Jhon Yrap Masuli Gahiton
      2 days ago




      $begingroup$
      i know the answer is 1/12 but the one I found uses theorem we havent tackled it yet in our class
      $endgroup$
      – Jhon Yrap Masuli Gahiton
      2 days ago




      3




      3




      $begingroup$
      $frac{a-b}{a^3-b^3} = frac{a-b}{(a-b)(a^2+ab+b^2)} = frac{1}{a^2+ab+b^2}$.
      $endgroup$
      – Toby Mak
      2 days ago






      $begingroup$
      $frac{a-b}{a^3-b^3} = frac{a-b}{(a-b)(a^2+ab+b^2)} = frac{1}{a^2+ab+b^2}$.
      $endgroup$
      – Toby Mak
      2 days ago













      6












      $begingroup$

      Hint$$a^3-b^3=(a-b)(a^2+ab+b^2)$$






      share|cite|improve this answer









      $endgroup$




















        6












        $begingroup$

        Hint$$a^3-b^3=(a-b)(a^2+ab+b^2)$$






        share|cite|improve this answer









        $endgroup$


















          6












          6








          6





          $begingroup$

          Hint$$a^3-b^3=(a-b)(a^2+ab+b^2)$$






          share|cite|improve this answer









          $endgroup$



          Hint$$a^3-b^3=(a-b)(a^2+ab+b^2)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          Mostafa AyazMostafa Ayaz

          18.9k3 gold badges10 silver badges43 bronze badges




          18.9k3 gold badges10 silver badges43 bronze badges


























              5












              $begingroup$

              From Peter Foreman's comment



              By the definition of a derivative:



              $$lim_{a to 8} frac{a^{1/3} - 8^{1/3}}{a-8}$$
              $$= frac{mathrm d}{mathrm da} left( a^{1/3} right) biggl|_{a=8}$$






              share|cite|improve this answer











              $endgroup$




















                5












                $begingroup$

                From Peter Foreman's comment



                By the definition of a derivative:



                $$lim_{a to 8} frac{a^{1/3} - 8^{1/3}}{a-8}$$
                $$= frac{mathrm d}{mathrm da} left( a^{1/3} right) biggl|_{a=8}$$






                share|cite|improve this answer











                $endgroup$


















                  5












                  5








                  5





                  $begingroup$

                  From Peter Foreman's comment



                  By the definition of a derivative:



                  $$lim_{a to 8} frac{a^{1/3} - 8^{1/3}}{a-8}$$
                  $$= frac{mathrm d}{mathrm da} left( a^{1/3} right) biggl|_{a=8}$$






                  share|cite|improve this answer











                  $endgroup$



                  From Peter Foreman's comment



                  By the definition of a derivative:



                  $$lim_{a to 8} frac{a^{1/3} - 8^{1/3}}{a-8}$$
                  $$= frac{mathrm d}{mathrm da} left( a^{1/3} right) biggl|_{a=8}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 2 days ago

























                  answered 2 days ago









                  Toby MakToby Mak

                  4,7451 gold badge14 silver badges29 bronze badges




                  4,7451 gold badge14 silver badges29 bronze badges


























                      4












                      $begingroup$

                      Recall that there are two equivalent definitions of the derivative of a function in calculus. We're interested in this one:



                      $$f'(a)=lim_{xto a}frac{f(x)-f(a)}{x-a}.$$



                      Your limit can be easily rewritten to fit that form. The only other thing that you would need here is the derivative of $sqrt[3]{x}$. It's quite easy to find using the so-called power rule. It's equal to $frac{1}{3sqrt[3]{x^2}}$. So, your problem now becomes the problem of evaluating the derivative of the function $sqrt[3]{x}$ at $x=8$:



                      $$
                      lim_{xto 8}frac{sqrt[3]{x}-2}{x-8}=
                      lim_{xto 8}frac{sqrt[3]{x}-sqrt[3]{8}}{x-8}=
                      frac{d}{dx}left(sqrt[3]{x}right)bigg|_{x=8}=\
                      frac{1}{3sqrt[3]{x^2}}bigg|_{x=8}=
                      frac{1}{3sqrt[3]{8^2}}=
                      frac{1}{3cdot 4}=
                      frac{1}{12}.
                      $$






                      share|cite|improve this answer











                      $endgroup$











                      • 1




                        $begingroup$
                        this is great but we are not yet on derivatives please without using derivatives I cant understand yet
                        $endgroup$
                        – Jhon Yrap Masuli Gahiton
                        2 days ago
















                      4












                      $begingroup$

                      Recall that there are two equivalent definitions of the derivative of a function in calculus. We're interested in this one:



                      $$f'(a)=lim_{xto a}frac{f(x)-f(a)}{x-a}.$$



                      Your limit can be easily rewritten to fit that form. The only other thing that you would need here is the derivative of $sqrt[3]{x}$. It's quite easy to find using the so-called power rule. It's equal to $frac{1}{3sqrt[3]{x^2}}$. So, your problem now becomes the problem of evaluating the derivative of the function $sqrt[3]{x}$ at $x=8$:



                      $$
                      lim_{xto 8}frac{sqrt[3]{x}-2}{x-8}=
                      lim_{xto 8}frac{sqrt[3]{x}-sqrt[3]{8}}{x-8}=
                      frac{d}{dx}left(sqrt[3]{x}right)bigg|_{x=8}=\
                      frac{1}{3sqrt[3]{x^2}}bigg|_{x=8}=
                      frac{1}{3sqrt[3]{8^2}}=
                      frac{1}{3cdot 4}=
                      frac{1}{12}.
                      $$






                      share|cite|improve this answer











                      $endgroup$











                      • 1




                        $begingroup$
                        this is great but we are not yet on derivatives please without using derivatives I cant understand yet
                        $endgroup$
                        – Jhon Yrap Masuli Gahiton
                        2 days ago














                      4












                      4








                      4





                      $begingroup$

                      Recall that there are two equivalent definitions of the derivative of a function in calculus. We're interested in this one:



                      $$f'(a)=lim_{xto a}frac{f(x)-f(a)}{x-a}.$$



                      Your limit can be easily rewritten to fit that form. The only other thing that you would need here is the derivative of $sqrt[3]{x}$. It's quite easy to find using the so-called power rule. It's equal to $frac{1}{3sqrt[3]{x^2}}$. So, your problem now becomes the problem of evaluating the derivative of the function $sqrt[3]{x}$ at $x=8$:



                      $$
                      lim_{xto 8}frac{sqrt[3]{x}-2}{x-8}=
                      lim_{xto 8}frac{sqrt[3]{x}-sqrt[3]{8}}{x-8}=
                      frac{d}{dx}left(sqrt[3]{x}right)bigg|_{x=8}=\
                      frac{1}{3sqrt[3]{x^2}}bigg|_{x=8}=
                      frac{1}{3sqrt[3]{8^2}}=
                      frac{1}{3cdot 4}=
                      frac{1}{12}.
                      $$






                      share|cite|improve this answer











                      $endgroup$



                      Recall that there are two equivalent definitions of the derivative of a function in calculus. We're interested in this one:



                      $$f'(a)=lim_{xto a}frac{f(x)-f(a)}{x-a}.$$



                      Your limit can be easily rewritten to fit that form. The only other thing that you would need here is the derivative of $sqrt[3]{x}$. It's quite easy to find using the so-called power rule. It's equal to $frac{1}{3sqrt[3]{x^2}}$. So, your problem now becomes the problem of evaluating the derivative of the function $sqrt[3]{x}$ at $x=8$:



                      $$
                      lim_{xto 8}frac{sqrt[3]{x}-2}{x-8}=
                      lim_{xto 8}frac{sqrt[3]{x}-sqrt[3]{8}}{x-8}=
                      frac{d}{dx}left(sqrt[3]{x}right)bigg|_{x=8}=\
                      frac{1}{3sqrt[3]{x^2}}bigg|_{x=8}=
                      frac{1}{3sqrt[3]{8^2}}=
                      frac{1}{3cdot 4}=
                      frac{1}{12}.
                      $$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 2 days ago

























                      answered 2 days ago









                      Michael RybkinMichael Rybkin

                      6,2542 gold badges6 silver badges26 bronze badges




                      6,2542 gold badges6 silver badges26 bronze badges











                      • 1




                        $begingroup$
                        this is great but we are not yet on derivatives please without using derivatives I cant understand yet
                        $endgroup$
                        – Jhon Yrap Masuli Gahiton
                        2 days ago














                      • 1




                        $begingroup$
                        this is great but we are not yet on derivatives please without using derivatives I cant understand yet
                        $endgroup$
                        – Jhon Yrap Masuli Gahiton
                        2 days ago








                      1




                      1




                      $begingroup$
                      this is great but we are not yet on derivatives please without using derivatives I cant understand yet
                      $endgroup$
                      – Jhon Yrap Masuli Gahiton
                      2 days ago




                      $begingroup$
                      this is great but we are not yet on derivatives please without using derivatives I cant understand yet
                      $endgroup$
                      – Jhon Yrap Masuli Gahiton
                      2 days ago











                      0












                      $begingroup$

                      There is a standard binomial limit which you may use
                      Limit x tend to a (x^n-a^n)/(x-a) = na^(n-1)
                      This is proved by using the maclaurin series expansions as far I as can recall...the answer by this method for your problem would be (1/12)






                      share|cite|improve this answer








                      New contributor



                      Schwarz Kugelblitz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.





                      $endgroup$















                      • $begingroup$
                        This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
                        $endgroup$
                        – Klangen
                        2 days ago










                      • $begingroup$
                        I fail to see how this doesn't answer the question.... this is a standard formula taught in India for competitive entrance exams like the JEE (joint entrance exam) with a decent amount of proofs behind it and a good number of authors publishing it in their books...the person who asked the question wanted an alternate approach other than l'hopital and I gave a valid mathematically sound one
                        $endgroup$
                        – Schwarz Kugelblitz
                        2 days ago






                      • 2




                        $begingroup$
                        Two things that would make this look more like an answer: give a reason for the formula that is better than "it's in a bunch of test prep books" (a proof of the formula would be such a reason), and format the formula in standard notation so that people can more easily recognize it: see math.meta.stackexchange.com/questions/5020/…
                        $endgroup$
                        – David K
                        2 days ago












                      • $begingroup$
                        What you would have to do is to show that $frac{x^n-a^n}{x-a} = x^n+x^{n-1}a + cdots + xa^{n-1} + a^n$, and then substitute $x=a$ to get to your answer. Bear in mind that this question might also be useful to others, who might not take JEE and read the books you are suggesting.
                        $endgroup$
                        – Toby Mak
                        2 days ago


















                      0












                      $begingroup$

                      There is a standard binomial limit which you may use
                      Limit x tend to a (x^n-a^n)/(x-a) = na^(n-1)
                      This is proved by using the maclaurin series expansions as far I as can recall...the answer by this method for your problem would be (1/12)






                      share|cite|improve this answer








                      New contributor



                      Schwarz Kugelblitz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.





                      $endgroup$















                      • $begingroup$
                        This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
                        $endgroup$
                        – Klangen
                        2 days ago










                      • $begingroup$
                        I fail to see how this doesn't answer the question.... this is a standard formula taught in India for competitive entrance exams like the JEE (joint entrance exam) with a decent amount of proofs behind it and a good number of authors publishing it in their books...the person who asked the question wanted an alternate approach other than l'hopital and I gave a valid mathematically sound one
                        $endgroup$
                        – Schwarz Kugelblitz
                        2 days ago






                      • 2




                        $begingroup$
                        Two things that would make this look more like an answer: give a reason for the formula that is better than "it's in a bunch of test prep books" (a proof of the formula would be such a reason), and format the formula in standard notation so that people can more easily recognize it: see math.meta.stackexchange.com/questions/5020/…
                        $endgroup$
                        – David K
                        2 days ago












                      • $begingroup$
                        What you would have to do is to show that $frac{x^n-a^n}{x-a} = x^n+x^{n-1}a + cdots + xa^{n-1} + a^n$, and then substitute $x=a$ to get to your answer. Bear in mind that this question might also be useful to others, who might not take JEE and read the books you are suggesting.
                        $endgroup$
                        – Toby Mak
                        2 days ago
















                      0












                      0








                      0





                      $begingroup$

                      There is a standard binomial limit which you may use
                      Limit x tend to a (x^n-a^n)/(x-a) = na^(n-1)
                      This is proved by using the maclaurin series expansions as far I as can recall...the answer by this method for your problem would be (1/12)






                      share|cite|improve this answer








                      New contributor



                      Schwarz Kugelblitz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.





                      $endgroup$



                      There is a standard binomial limit which you may use
                      Limit x tend to a (x^n-a^n)/(x-a) = na^(n-1)
                      This is proved by using the maclaurin series expansions as far I as can recall...the answer by this method for your problem would be (1/12)







                      share|cite|improve this answer








                      New contributor



                      Schwarz Kugelblitz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.








                      share|cite|improve this answer



                      share|cite|improve this answer






                      New contributor



                      Schwarz Kugelblitz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.








                      answered 2 days ago









                      Schwarz KugelblitzSchwarz Kugelblitz

                      61 bronze badge




                      61 bronze badge




                      New contributor



                      Schwarz Kugelblitz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.




                      New contributor




                      Schwarz Kugelblitz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.

















                      • $begingroup$
                        This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
                        $endgroup$
                        – Klangen
                        2 days ago










                      • $begingroup$
                        I fail to see how this doesn't answer the question.... this is a standard formula taught in India for competitive entrance exams like the JEE (joint entrance exam) with a decent amount of proofs behind it and a good number of authors publishing it in their books...the person who asked the question wanted an alternate approach other than l'hopital and I gave a valid mathematically sound one
                        $endgroup$
                        – Schwarz Kugelblitz
                        2 days ago






                      • 2




                        $begingroup$
                        Two things that would make this look more like an answer: give a reason for the formula that is better than "it's in a bunch of test prep books" (a proof of the formula would be such a reason), and format the formula in standard notation so that people can more easily recognize it: see math.meta.stackexchange.com/questions/5020/…
                        $endgroup$
                        – David K
                        2 days ago












                      • $begingroup$
                        What you would have to do is to show that $frac{x^n-a^n}{x-a} = x^n+x^{n-1}a + cdots + xa^{n-1} + a^n$, and then substitute $x=a$ to get to your answer. Bear in mind that this question might also be useful to others, who might not take JEE and read the books you are suggesting.
                        $endgroup$
                        – Toby Mak
                        2 days ago




















                      • $begingroup$
                        This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
                        $endgroup$
                        – Klangen
                        2 days ago










                      • $begingroup$
                        I fail to see how this doesn't answer the question.... this is a standard formula taught in India for competitive entrance exams like the JEE (joint entrance exam) with a decent amount of proofs behind it and a good number of authors publishing it in their books...the person who asked the question wanted an alternate approach other than l'hopital and I gave a valid mathematically sound one
                        $endgroup$
                        – Schwarz Kugelblitz
                        2 days ago






                      • 2




                        $begingroup$
                        Two things that would make this look more like an answer: give a reason for the formula that is better than "it's in a bunch of test prep books" (a proof of the formula would be such a reason), and format the formula in standard notation so that people can more easily recognize it: see math.meta.stackexchange.com/questions/5020/…
                        $endgroup$
                        – David K
                        2 days ago












                      • $begingroup$
                        What you would have to do is to show that $frac{x^n-a^n}{x-a} = x^n+x^{n-1}a + cdots + xa^{n-1} + a^n$, and then substitute $x=a$ to get to your answer. Bear in mind that this question might also be useful to others, who might not take JEE and read the books you are suggesting.
                        $endgroup$
                        – Toby Mak
                        2 days ago


















                      $begingroup$
                      This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
                      $endgroup$
                      – Klangen
                      2 days ago




                      $begingroup$
                      This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
                      $endgroup$
                      – Klangen
                      2 days ago












                      $begingroup$
                      I fail to see how this doesn't answer the question.... this is a standard formula taught in India for competitive entrance exams like the JEE (joint entrance exam) with a decent amount of proofs behind it and a good number of authors publishing it in their books...the person who asked the question wanted an alternate approach other than l'hopital and I gave a valid mathematically sound one
                      $endgroup$
                      – Schwarz Kugelblitz
                      2 days ago




                      $begingroup$
                      I fail to see how this doesn't answer the question.... this is a standard formula taught in India for competitive entrance exams like the JEE (joint entrance exam) with a decent amount of proofs behind it and a good number of authors publishing it in their books...the person who asked the question wanted an alternate approach other than l'hopital and I gave a valid mathematically sound one
                      $endgroup$
                      – Schwarz Kugelblitz
                      2 days ago




                      2




                      2




                      $begingroup$
                      Two things that would make this look more like an answer: give a reason for the formula that is better than "it's in a bunch of test prep books" (a proof of the formula would be such a reason), and format the formula in standard notation so that people can more easily recognize it: see math.meta.stackexchange.com/questions/5020/…
                      $endgroup$
                      – David K
                      2 days ago






                      $begingroup$
                      Two things that would make this look more like an answer: give a reason for the formula that is better than "it's in a bunch of test prep books" (a proof of the formula would be such a reason), and format the formula in standard notation so that people can more easily recognize it: see math.meta.stackexchange.com/questions/5020/…
                      $endgroup$
                      – David K
                      2 days ago














                      $begingroup$
                      What you would have to do is to show that $frac{x^n-a^n}{x-a} = x^n+x^{n-1}a + cdots + xa^{n-1} + a^n$, and then substitute $x=a$ to get to your answer. Bear in mind that this question might also be useful to others, who might not take JEE and read the books you are suggesting.
                      $endgroup$
                      – Toby Mak
                      2 days ago






                      $begingroup$
                      What you would have to do is to show that $frac{x^n-a^n}{x-a} = x^n+x^{n-1}a + cdots + xa^{n-1} + a^n$, and then substitute $x=a$ to get to your answer. Bear in mind that this question might also be useful to others, who might not take JEE and read the books you are suggesting.
                      $endgroup$
                      – Toby Mak
                      2 days ago





                      -

                      Popular posts from this blog

                      Taj Mahal Inhaltsverzeichnis Aufbau | Geschichte | 350-Jahr-Feier | Heutige Bedeutung | Siehe auch |...

                      Image
                      Mogul-ArchitekturMausoleum in IndienBauwerk in Uttar PradeshGrabmoscheeKuppelbauwerkWeltkulturerbe in AsienWeltkulturerbe in IndienAgraMogulreichGrabbau in AsienGrabgartenErbaut in den 1640er Jahren deutschpersoarabischDMGDevanagariMausoleumYamunaAgraindischenUttar PradeshKuppelMinaretteZiegelsteinenMarmorplattenGroßmogulShah JahanMumtaz MahalOktogonTambourgebauchtenChhatriIwanMinaretteErdbebenWendeltreppeJaliKenotapheKryptaReliefsMarmorinkrustationenKoranMärtyrerinGottesSandsteinTschahār BāghBagh-e WafaDschalalabadParadiesgartenMekkaMoscheeSüd-ZentralasiensAhmad LahoriBadakhshanAbu Fazelpersische Architekturindo-islamischen BaukunstIndienAsiensElefantenMarmorJaipurRajasthanNarbadaEdelsteinenHalbedelsteinenJaspisPunjabKarneoleAchatYemenKristallChinaLapislazuliSaphireCeylonKorallenArabienRoten MeerGranateBundelkhandDiamantenPannaOnyxPersienHeliotropFatehpur SikriTürkiseJean-Baptiste TavernierFrançois BernierMullahsSeelenTeppichenYamunaMuhammad Aurangzeb AlamgirTschahār BāghHumayunsbri...
                      Read more

                      Baia Sprie Cuprins Etimologie | Istorie | Demografie | Politică și administrație | Arii naturale...

                      Image
                      Localități urbane în județul MaramureșOrașe în județul MaramureșBaia Sprie maghiarăgermanăorașjudețul MaramureșTransilvaniaRomâniaChiuzbaiaTăuții de SusSatu Nou de Susminieresașilor13291329maghiarCarol RobertBaia Mare13601348139014061851latinescul1540germanBaia Mare1564Nicolae OlahusSăsarGutâiBaia Mare1411Belgradului1427Iancu de Hunedoara14461453Matei CorvinMaria Terezaslovacirecensământului efectuat în 2011recensământul anterior din 2002românimaghiariromiortodocșiromano-catolicireformațigreco-catolicipenticostalimartori ai lui IehovaSebastian Alin BârdaPartidul Național Liberal178019321960școlii băimărene de picturăAurel PoppAdy EndreNicolae TonitzaAstra19221997România1876vulcanicetectonicădaciteandezitecuarțuluiBaia SprieBorșaCavnicDragomireștiSăliștea de SusSeiniȘomcuta MareTârgu LăpușTăuții-MăgherăușUlmeniVișeu de SusArdusatArinișAsuaju de SusBăița de sub CodruBăiuțBârsanaBăseștiBicazBistraBocicoiu MareBogdan VodăBoiu MareBotizaBudeștiCălineștiCâmpulung la TisaCerneștiCicârlăuCoa...
                      Read more

                      Nicolae Petrescu-Găină Cuprins Biografie | Opera | In memoriam | Varia | Controverse, incertitudini...

                      Image
                      Articole de calitateNașteri în 1871Nașteri pe 31 martieDecese în 1931Decese pe 15 februarieArtiști români din secolul al XIX-leaArtiști români din secolul al XX-leaCaricaturiști româniTinerimea artistică CraiovaRomâniaBucureștiRomâniacaricaturistȘtefan LuchianConstantin ArtachinoAlexandru Bogdan-PiteștiNicolae Vermont1898Honoré DaumierParlamentului României19081913Constantin Jiquidi1898secolului al XIX-leaGolemo KonjariBulgariaMuzeului de Artă din Craiova2007Craiova31 martie1871limba francezălimba latinăLiceul Sfântul SavaBucureștigazului metanOperei RomâneȘtefan Petrescu-MuscăLiceul „Carol I”Școlii de Belle-ArteCostin PetrescuGheorghe TattarescuTheodor AmanConstantin I. StăncescuAlexandru ObedenaruSalonului Artiștilor IndependențiExpoziția Artiștilor în ViațăNicolae VermontȘtefan LuchianConstantin ArtachinoAlexandru Bogdan-PiteștiIon AngelescuNicolae GrantIon TicuLudovic DolinskiCaricatura Eu Sunt... Școala de Belle-Arte din BucureștiAteneului RomânGeorge RanettiCafeneaua ImperialFi...
                      Read more