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...
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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?
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$begingroup$
$$lim_{xto8}frac{sqrt[3]x-2}{x-8}$$
Is there any way I will not use l'Hôpital's rule here?
calculus limits limits-without-lhopital
edited 2 days ago
Asaf Karagila♦
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Jhon Yrap Masuli GahitonJhon Yrap Masuli Gahiton
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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:
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If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$$lim_{xto8}frac{sqrt[3]x-2}{x-8}$$
Is there any way I will not use l'Hôpital's rule here?
calculus limits limits-without-lhopital
edited 2 days ago
Asaf Karagila♦
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asked 2 days ago
Jhon Yrap Masuli GahitonJhon Yrap Masuli Gahiton
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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
add a comment |
$begingroup$
$$lim_{xto8}frac{sqrt[3]x-2}{x-8}$$
Is there any way I will not use l'Hôpital's rule here?
calculus limits limits-without-lhopital
edited 2 days ago
Asaf Karagila♦
315k35 gold badges454 silver badges788 bronze badges
asked 2 days ago
Jhon Yrap Masuli GahitonJhon Yrap Masuli Gahiton
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.
$endgroup$
$$lim_{xto8}frac{sqrt[3]x-2}{x-8}$$
Is there any way I will not use l'Hôpital's rule here?
calculus limits limits-without-lhopital
calculus limits limits-without-lhopital
edited 2 days ago
Asaf Karagila♦
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asked 2 days ago
Jhon Yrap Masuli GahitonJhon Yrap Masuli Gahiton
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edited 2 days ago
Asaf Karagila♦
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asked 2 days ago
Jhon Yrap Masuli GahitonJhon Yrap Masuli Gahiton
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edited 2 days ago
Asaf Karagila♦
315k35 gold badges454 silver badges788 bronze badges
edited 2 days ago
Asaf Karagila♦
315k35 gold badges454 silver badges788 bronze badges
edited 2 days ago
Asaf Karagila♦
315k35 gold badges454 silver badges788 bronze badges
315k35 gold badges454 silver badges788 bronze badges
asked 2 days ago
Jhon Yrap Masuli GahitonJhon Yrap Masuli Gahiton
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asked 2 days ago
Jhon Yrap Masuli GahitonJhon Yrap Masuli Gahiton
193 bronze badges
asked 2 days ago
Jhon Yrap Masuli GahitonJhon Yrap Masuli Gahiton
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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:
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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
add a comment |
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
add a comment |
6 Answers
6
active
oldest
votes
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The denominator can be factored as
$$(x^{1/3}-2)(x^{2/3}+2x^{1/3}+4)$$
so
$$lim_{xto8}frac{x^{1/3}-2}{x-8}=lim_{xto8}frac1{x^{2/3}+2x^{1/3}+4}=frac1{12}$$
answered 2 days ago
Parcly TaxelParcly Taxel
51.5k13 gold badges80 silver badges120 bronze badges
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Hint.
$$
frac{sqrt[3]{x}-2}{x-8} = frac{sqrt[3]{x}-2}{(sqrt[3]{x})^3-2^3}
$$
answered 2 days ago
CesareoCesareo
12.4k3 gold badges6 silver badges21 bronze badges
$endgroup$
$begingroup$
can you show me the steps hehe badly needed help
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– 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
add a comment |
$begingroup$
Hint$$a^3-b^3=(a-b)(a^2+ab+b^2)$$
answered 2 days ago
Mostafa AyazMostafa Ayaz
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add a comment |
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From Peter Foreman's comment
By the definition of a derivative:
$$lim_{a to 8} frac{a^{1/3} - 8^{1/3}}{a-8}$$
$$= frac{mathrm d}{mathrm da} left( a^{1/3} right) biggl|_{a=8}$$
answered 2 days ago
Toby MakToby Mak
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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 $x=8$:
$$
lim_{xto 8}frac{sqrt[3]{x}-2}{x-8}=
lim_{xto 8}frac{sqrt[3]{x}-sqrt[3]{8}}{x-8}=
frac{d}{dx}left(sqrt[3]{x}right)bigg|_{x=8}=\
frac{1}{3sqrt[3]{x^2}}bigg|_{x=8}=
frac{1}{3sqrt[3]{8^2}}=
frac{1}{3cdot 4}=
frac{1}{12}.
$$
answered 2 days ago
Michael RybkinMichael Rybkin
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1
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this is great but we are not yet on derivatives please without using derivatives I cant understand yet
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– Jhon Yrap Masuli Gahiton
2 days ago
add a comment |
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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)
answered 2 days ago
Schwarz KugelblitzSchwarz Kugelblitz
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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
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– Klangen
2 days ago
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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
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– 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/…
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– 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
add a comment |
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The denominator can be factored as
$$(x^{1/3}-2)(x^{2/3}+2x^{1/3}+4)$$
so
$$lim_{xto8}frac{x^{1/3}-2}{x-8}=lim_{xto8}frac1{x^{2/3}+2x^{1/3}+4}=frac1{12}$$
answered 2 days ago
Parcly TaxelParcly Taxel
51.5k13 gold badges80 silver badges120 bronze badges
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
The denominator can be factored as
$$(x^{1/3}-2)(x^{2/3}+2x^{1/3}+4)$$
so
$$lim_{xto8}frac{x^{1/3}-2}{x-8}=lim_{xto8}frac1{x^{2/3}+2x^{1/3}+4}=frac1{12}$$
answered 2 days ago
Parcly TaxelParcly Taxel
51.5k13 gold badges80 silver badges120 bronze badges
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
The denominator can be factored as
$$(x^{1/3}-2)(x^{2/3}+2x^{1/3}+4)$$
so
$$lim_{xto8}frac{x^{1/3}-2}{x-8}=lim_{xto8}frac1{x^{2/3}+2x^{1/3}+4}=frac1{12}$$
answered 2 days ago
Parcly TaxelParcly Taxel
51.5k13 gold badges80 silver badges120 bronze badges
$endgroup$
The denominator can be factored as
$$(x^{1/3}-2)(x^{2/3}+2x^{1/3}+4)$$
so
$$lim_{xto8}frac{x^{1/3}-2}{x-8}=lim_{xto8}frac1{x^{2/3}+2x^{1/3}+4}=frac1{12}$$
answered 2 days ago
Parcly TaxelParcly Taxel
51.5k13 gold badges80 silver badges120 bronze badges
answered 2 days ago
Parcly TaxelParcly Taxel
51.5k13 gold badges80 silver badges120 bronze badges
answered 2 days ago
Parcly TaxelParcly Taxel
51.5k13 gold badges80 silver badges120 bronze badges
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
Hint.
$$
frac{sqrt[3]{x}-2}{x-8} = frac{sqrt[3]{x}-2}{(sqrt[3]{x})^3-2^3}
$$
answered 2 days ago
CesareoCesareo
12.4k3 gold badges6 silver badges21 bronze badges
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Hint.
$$
frac{sqrt[3]{x}-2}{x-8} = frac{sqrt[3]{x}-2}{(sqrt[3]{x})^3-2^3}
$$
answered 2 days ago
CesareoCesareo
12.4k3 gold badges6 silver badges21 bronze badges
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Hint.
$$
frac{sqrt[3]{x}-2}{x-8} = frac{sqrt[3]{x}-2}{(sqrt[3]{x})^3-2^3}
$$
answered 2 days ago
CesareoCesareo
12.4k3 gold badges6 silver badges21 bronze badges
$endgroup$
Hint.
$$
frac{sqrt[3]{x}-2}{x-8} = frac{sqrt[3]{x}-2}{(sqrt[3]{x})^3-2^3}
$$
answered 2 days ago
CesareoCesareo
12.4k3 gold badges6 silver badges21 bronze badges
answered 2 days ago
CesareoCesareo
12.4k3 gold badges6 silver badges21 bronze badges
answered 2 days ago
CesareoCesareo
12.4k3 gold badges6 silver badges21 bronze badges
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
add a comment |
$begingroup$
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
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– 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
$begingroup$
$frac{a-b}{a^3-b^3} = frac{a-b}{(a-b)(a^2+ab+b^2)} = frac{1}{a^2+ab+b^2}$.
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– 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
add a comment |
$begingroup$
Hint$$a^3-b^3=(a-b)(a^2+ab+b^2)$$
answered 2 days ago
Mostafa AyazMostafa Ayaz
18.9k3 gold badges10 silver badges43 bronze badges
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add a comment |
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Hint$$a^3-b^3=(a-b)(a^2+ab+b^2)$$
answered 2 days ago
Mostafa AyazMostafa Ayaz
18.9k3 gold badges10 silver badges43 bronze badges
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add a comment |
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Hint$$a^3-b^3=(a-b)(a^2+ab+b^2)$$
answered 2 days ago
Mostafa AyazMostafa Ayaz
18.9k3 gold badges10 silver badges43 bronze badges
$endgroup$
Hint$$a^3-b^3=(a-b)(a^2+ab+b^2)$$
answered 2 days ago
Mostafa AyazMostafa Ayaz
18.9k3 gold badges10 silver badges43 bronze badges
answered 2 days ago
Mostafa AyazMostafa Ayaz
18.9k3 gold badges10 silver badges43 bronze badges
answered 2 days ago
Mostafa AyazMostafa Ayaz
18.9k3 gold badges10 silver badges43 bronze badges
answered 2 days ago
Mostafa AyazMostafa Ayaz
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18.9k3 gold badges10 silver badges43 bronze badges
add a comment |
add a comment |
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From Peter Foreman's comment
By the definition of a derivative:
$$lim_{a to 8} frac{a^{1/3} - 8^{1/3}}{a-8}$$
$$= frac{mathrm d}{mathrm da} left( a^{1/3} right) biggl|_{a=8}$$
answered 2 days ago
Toby MakToby Mak
4,7451 gold badge14 silver badges29 bronze badges
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add a comment |
$begingroup$
From Peter Foreman's comment
By the definition of a derivative:
$$lim_{a to 8} frac{a^{1/3} - 8^{1/3}}{a-8}$$
$$= frac{mathrm d}{mathrm da} left( a^{1/3} right) biggl|_{a=8}$$
answered 2 days ago
Toby MakToby Mak
4,7451 gold badge14 silver badges29 bronze badges
$endgroup$
add a comment |
$begingroup$
From Peter Foreman's comment
By the definition of a derivative:
$$lim_{a to 8} frac{a^{1/3} - 8^{1/3}}{a-8}$$
$$= frac{mathrm d}{mathrm da} left( a^{1/3} right) biggl|_{a=8}$$
answered 2 days ago
Toby MakToby Mak
4,7451 gold badge14 silver badges29 bronze badges
$endgroup$
From Peter Foreman's comment
By the definition of a derivative:
$$lim_{a to 8} frac{a^{1/3} - 8^{1/3}}{a-8}$$
$$= frac{mathrm d}{mathrm da} left( a^{1/3} right) biggl|_{a=8}$$
answered 2 days ago
Toby MakToby Mak
4,7451 gold badge14 silver badges29 bronze badges
edited 2 days ago
answered 2 days ago
Toby MakToby Mak
4,7451 gold badge14 silver badges29 bronze badges
answered 2 days ago
Toby MakToby Mak
4,7451 gold badge14 silver badges29 bronze badges
answered 2 days ago
Toby MakToby Mak
4,7451 gold badge14 silver badges29 bronze badges
4,7451 gold badge14 silver badges29 bronze badges
add a comment |
add a comment |
$begingroup$
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 $x=8$:
$$
lim_{xto 8}frac{sqrt[3]{x}-2}{x-8}=
lim_{xto 8}frac{sqrt[3]{x}-sqrt[3]{8}}{x-8}=
frac{d}{dx}left(sqrt[3]{x}right)bigg|_{x=8}=\
frac{1}{3sqrt[3]{x^2}}bigg|_{x=8}=
frac{1}{3sqrt[3]{8^2}}=
frac{1}{3cdot 4}=
frac{1}{12}.
$$
answered 2 days ago
Michael RybkinMichael Rybkin
6,2542 gold badges6 silver badges26 bronze badges
$endgroup$
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
add a comment |
$begingroup$
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 $x=8$:
$$
lim_{xto 8}frac{sqrt[3]{x}-2}{x-8}=
lim_{xto 8}frac{sqrt[3]{x}-sqrt[3]{8}}{x-8}=
frac{d}{dx}left(sqrt[3]{x}right)bigg|_{x=8}=\
frac{1}{3sqrt[3]{x^2}}bigg|_{x=8}=
frac{1}{3sqrt[3]{8^2}}=
frac{1}{3cdot 4}=
frac{1}{12}.
$$
answered 2 days ago
Michael RybkinMichael Rybkin
6,2542 gold badges6 silver badges26 bronze badges
$endgroup$
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
add a comment |
$begingroup$
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 $x=8$:
$$
lim_{xto 8}frac{sqrt[3]{x}-2}{x-8}=
lim_{xto 8}frac{sqrt[3]{x}-sqrt[3]{8}}{x-8}=
frac{d}{dx}left(sqrt[3]{x}right)bigg|_{x=8}=\
frac{1}{3sqrt[3]{x^2}}bigg|_{x=8}=
frac{1}{3sqrt[3]{8^2}}=
frac{1}{3cdot 4}=
frac{1}{12}.
$$
answered 2 days ago
Michael RybkinMichael Rybkin
6,2542 gold badges6 silver badges26 bronze badges
$endgroup$
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 $x=8$:
$$
lim_{xto 8}frac{sqrt[3]{x}-2}{x-8}=
lim_{xto 8}frac{sqrt[3]{x}-sqrt[3]{8}}{x-8}=
frac{d}{dx}left(sqrt[3]{x}right)bigg|_{x=8}=\
frac{1}{3sqrt[3]{x^2}}bigg|_{x=8}=
frac{1}{3sqrt[3]{8^2}}=
frac{1}{3cdot 4}=
frac{1}{12}.
$$
answered 2 days ago
Michael RybkinMichael Rybkin
6,2542 gold badges6 silver badges26 bronze badges
edited 2 days ago
answered 2 days ago
Michael RybkinMichael Rybkin
6,2542 gold badges6 silver badges26 bronze badges
answered 2 days ago
Michael RybkinMichael Rybkin
6,2542 gold badges6 silver badges26 bronze badges
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
add a comment |
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
add a comment |
$begingroup$
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)
answered 2 days ago
Schwarz KugelblitzSchwarz Kugelblitz
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.
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$begingroup$
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
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– 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/…
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– 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
add a comment |
$begingroup$
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)
answered 2 days ago
Schwarz KugelblitzSchwarz Kugelblitz
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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)
answered 2 days ago
Schwarz KugelblitzSchwarz Kugelblitz
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.
$endgroup$
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)
answered 2 days ago
Schwarz KugelblitzSchwarz Kugelblitz
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.
answered 2 days ago
Schwarz KugelblitzSchwarz Kugelblitz
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.
answered 2 days ago
Schwarz KugelblitzSchwarz Kugelblitz
61 bronze badge
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
add a comment |
$begingroup$
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
add a comment |
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?
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– Toby Mak
2 days ago
$begingroup$
How do you suppose we define $f'(8)$ where $f(x)=sqrt[3]{x}$?
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– Peter Foreman
2 days ago