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Value of a limit.

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Value of a limit.

Using Basics Limit ArithmeticsLimits problem to find the values of constants - a and b If $lim_{x to infty}(1+frac{a}{x}+frac{b}{x^2})^{2x}=e^2$ Find the value of $a$ and $b$.The limit of $sin(n^alpha)$finding limit: result division by nullWhat is the value of $lfloor{100N}rfloor$Solve limit with Lagrange theoremA limit of n times sine values at factorially spaced argumentslimit and derivative questionDetermine this limit using L'Hopitals ruleApplication of Cauchy's first limit theorem

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2

$begingroup$

The value of the $$limlimits_{xto-infty}{(4x^2-x)^{1/2} +2x}$$ is?

The answer given is $1/4$.

I rationalized and got $$limlimits_{xto-infty} frac {-x}{|x|[(4- frac {1}{x})^{1/2}-2]}$$ how to proceed further?

algebra-precalculus limits

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edited 9 hours ago

Tapi

asked 9 hours ago

TapiTapi

43211 silver badge1717 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YvesDaoust Yes, I misread it.
  $endgroup$
  – saulspatz
  9 hours ago
  

  
  
  
  


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  I don't see how your rationalization works. $-2|x|neq -2x.$ In particular, when $x$ is negative $2x=-2|x|.$
  $endgroup$
  – Thomas Andrews
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You should get a rationalization: $$frac{|x|}{|x|left[left(4-frac{1}{x}right)^{1/2}+2right]}$$ when $x<0.$
  $endgroup$
  – Thomas Andrews
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasAndrews Yes, that's where I made the mistake.
  $endgroup$
  – Tapi
  8 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

The value of the $$limlimits_{xto-infty}{(4x^2-x)^{1/2} +2x}$$ is?

The answer given is $1/4$.

I rationalized and got $$limlimits_{xto-infty} frac {-x}{|x|[(4- frac {1}{x})^{1/2}-2]}$$ how to proceed further?

algebra-precalculus limits

share|cite|improve this question

edited 9 hours ago

Tapi

asked 9 hours ago

TapiTapi

43211 silver badge1717 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YvesDaoust Yes, I misread it.
  $endgroup$
  – saulspatz
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don't see how your rationalization works. $-2|x|neq -2x.$ In particular, when $x$ is negative $2x=-2|x|.$
  $endgroup$
  – Thomas Andrews
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You should get a rationalization: $$frac{|x|}{|x|left[left(4-frac{1}{x}right)^{1/2}+2right]}$$ when $x<0.$
  $endgroup$
  – Thomas Andrews
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ThomasAndrews Yes, that's where I made the mistake.
  $endgroup$
  – Tapi
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

The value of the $$limlimits_{xto-infty}{(4x^2-x)^{1/2} +2x}$$ is?

The answer given is $1/4$.

I rationalized and got $$limlimits_{xto-infty} frac {-x}{|x|[(4- frac {1}{x})^{1/2}-2]}$$ how to proceed further?

algebra-precalculus limits

share|cite|improve this question

edited 9 hours ago

Tapi

asked 9 hours ago

TapiTapi

43211 silver badge1717 bronze badges

$endgroup$

share|cite|improve this question

edited 9 hours ago

Tapi

asked 9 hours ago

TapiTapi

43211 silver badge1717 bronze badges

share|cite|improve this question

edited 9 hours ago

Tapi

asked 9 hours ago

TapiTapi

43211 silver badge1717 bronze badges

asked 9 hours ago

TapiTapi

43211 silver badge1717 bronze badges

asked 9 hours ago

TapiTapi

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7 Answers
7

active

oldest

votes

3

$begingroup$

The limit is equivalent to
$$begin{align}
lim_{xto-infty}left(2|x|left(1-frac1{4x}right)^{1/2}+2xright)
&=lim_{xto-infty}left(-2xleft(1-frac1{4x}right)^{1/2}+2xright)\
&=lim_{xto-infty}-2xleft(left(1-frac1{4x}right)^{1/2}-1right)\
end{align}$$
Then using the generalized binomial expansion we get that as $xto0$
$$(1+x)^n=1+nx+o(x)$$
Hence our limit becomes
$$begin{align}
lim_{xto-infty}-2xleft(left(1-frac1{4x}right)^{1/2}-1right)
&=lim_{xto-infty}-2xleft(1-frac1{8x}+oleft(frac1xright)-1right)\
&=lim_{xto-infty}-2xleft(-frac1{8x}+oleft(frac1xright)right)\
&=lim_{xto-infty}left(frac14+o(1)right)\
&=frac14\
end{align}$$

share|cite|improve this answer

answered 8 hours ago

Peter ForemanPeter Foreman

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4

$begingroup$

A more elemental solution :
$$lim_{xto -infty} big( sqrt{4x^2-x} +2x big) =lim_{xto -infty} frac{-x}{sqrt{4x^2-x} -2x} =lim_{xto infty} frac{x}{sqrt{4x^2+x}+2x} =lim_{xtoinfty} frac{1}{sqrt{4+frac{1}{x}}+2} = frac{1}{sqrt{4}+2}=frac{1}{4}$$
where I used that
$$lim_{xto infty} f(x)=lim_{xto -infty} f(-x)$$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Azif00Azif00

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add a comment | 

1

$begingroup$

begin{align*}
(4x^2-x)^{1/2}+2x
&=sqrt{4x^2-x}+2x
\&=frac{(sqrt{4x^2-x}+2x)(sqrt{4x^2-x}-2x)}{sqrt{4x^2-x}-2x}
\&=frac{(4x^2-x)-4x^2}{sqrt{4x^2-x}-2x}
\&=frac{4x^2-x-4x^2}{sqrt{4x^2-x}-2x}
\&=frac{-x}{sqrt{4x^2-x}-2x}
\&=frac{-x}{sqrt{(2x)^2(1-frac{1}{4x})}-2x}
\&=frac{-x}{2|x|sqrt{1-frac{1}{4x}}-2x}
\&qquad [x<0]
\&=frac{-x}{-2xsqrt{1-frac{1}{4x}}-2x}
\&=frac{1}{2sqrt{1-frac{1}{4x}}+2}
\&tofrac{1}{2sqrt{1+0}+2}
\&=frac{1}{4}.
end{align*}
(as $xto-infty$)

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answered 8 hours ago

mf67mf67

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1

$begingroup$

Your rationalization is almost correct. It should be

$$frac{-x}{sqrt{4x^2-x}-2x} = frac{-x}{|x|left[left(4-frac{1}{x}right)^{1/2}-frac{2x}{|x|}right]}.$$

You have assumed that $-2x = -2|x|,$ which is not true when $x$ is negative. But as $xto-infty,$ we can asume $x<0$ and thus $|x|=-x,$ so this is equal to

$$frac{1}{left(4-frac1xright)^{1/2}+2}.$$

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edited 7 hours ago

Mars Plastic

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answered 8 hours ago

Thomas AndrewsThomas Andrews

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0

$begingroup$

For the sake of comfort, we change the sign and evaluate

$$limlimits_{xtoinfty}{(4x^2+x)^{1/2}-2x}$$

which is

$$limlimits_{xtoinfty}frac{x}{{(4x^2+x)^{1/2}+2x}}$$

or

$$limlimits_{xtoinfty}frac{1}{{(4+frac1x)^{1/2}+2}}=frac14.$$

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answered 8 hours ago

Yves DaoustYves Daoust

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0

$begingroup$

$y:=-x$, and $ lim y rightarrow +infty$.

$(4y^2+y)^{1/2}-2y=$

$((2y+1/4)^2-1/16)^{1/2}-2y$;

$z:=2y+1/4$;

We get

$((z^2-1/16)^{1/2}-z) +1/4$;

Since

$lim_{z rightarrow infty} (z^2-1/16)^{1/2}-z)=0$ (why?), and we are done.

Note:

$(z^2-1/16)^{1/2}-(z^2)^{1/2}= $

$dfrac{-1/16}{(z^2-1/16)^{1/2}+(z^2)^{1/2}}$

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edited 7 hours ago

answered 8 hours ago

Peter SzilasPeter Szilas

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0

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Hint: Multiply by ${sqrt{4x^2-x}-2x}over {sqrt{4x^2-x}-2x}$

The result is ${-xover{sqrt{4x^2-x}-2x}}$

$={{-x}over {|x|sqrt{4-{1over x}}+2}}=$

${{-x}over {-xsqrt{4-{1over x}}+2}}$

${1over {sqrt{4-{1over x}}+2}}$.

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edited 3 hours ago

answered 9 hours ago

Tsemo AristideTsemo Aristide

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $-2x$ is not under the root.
  $endgroup$
  – Tapi
  8 hours ago
  

  
  
  
  


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  $begingroup$
  $-2x=2|x|,$ at least when $x<0.$
  $endgroup$
  – Thomas Andrews
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This answer is still wrong. Instead of $-frac 2x$ in the denominator, it should be simple $+2.$
  $endgroup$
  – Thomas Andrews
  6 hours ago
  

  
  
  
  

add a comment | 

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3

$begingroup$

The limit is equivalent to
$$begin{align}
lim_{xto-infty}left(2|x|left(1-frac1{4x}right)^{1/2}+2xright)
&=lim_{xto-infty}left(-2xleft(1-frac1{4x}right)^{1/2}+2xright)\
&=lim_{xto-infty}-2xleft(left(1-frac1{4x}right)^{1/2}-1right)\
end{align}$$
Then using the generalized binomial expansion we get that as $xto0$
$$(1+x)^n=1+nx+o(x)$$
Hence our limit becomes
$$begin{align}
lim_{xto-infty}-2xleft(left(1-frac1{4x}right)^{1/2}-1right)
&=lim_{xto-infty}-2xleft(1-frac1{8x}+oleft(frac1xright)-1right)\
&=lim_{xto-infty}-2xleft(-frac1{8x}+oleft(frac1xright)right)\
&=lim_{xto-infty}left(frac14+o(1)right)\
&=frac14\
end{align}$$

share|cite|improve this answer

answered 8 hours ago

Peter ForemanPeter Foreman

13k11 gold badge55 silver badges2929 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

The limit is equivalent to
$$begin{align}
lim_{xto-infty}left(2|x|left(1-frac1{4x}right)^{1/2}+2xright)
&=lim_{xto-infty}left(-2xleft(1-frac1{4x}right)^{1/2}+2xright)\
&=lim_{xto-infty}-2xleft(left(1-frac1{4x}right)^{1/2}-1right)\
end{align}$$
Then using the generalized binomial expansion we get that as $xto0$
$$(1+x)^n=1+nx+o(x)$$
Hence our limit becomes
$$begin{align}
lim_{xto-infty}-2xleft(left(1-frac1{4x}right)^{1/2}-1right)
&=lim_{xto-infty}-2xleft(1-frac1{8x}+oleft(frac1xright)-1right)\
&=lim_{xto-infty}-2xleft(-frac1{8x}+oleft(frac1xright)right)\
&=lim_{xto-infty}left(frac14+o(1)right)\
&=frac14\
end{align}$$

share|cite|improve this answer

answered 8 hours ago

Peter ForemanPeter Foreman

13k11 gold badge55 silver badges2929 bronze badges

$endgroup$

add a comment | 

$begingroup$

The limit is equivalent to
$$begin{align}
lim_{xto-infty}left(2|x|left(1-frac1{4x}right)^{1/2}+2xright)
&=lim_{xto-infty}left(-2xleft(1-frac1{4x}right)^{1/2}+2xright)\
&=lim_{xto-infty}-2xleft(left(1-frac1{4x}right)^{1/2}-1right)\
end{align}$$
Then using the generalized binomial expansion we get that as $xto0$
$$(1+x)^n=1+nx+o(x)$$
Hence our limit becomes
$$begin{align}
lim_{xto-infty}-2xleft(left(1-frac1{4x}right)^{1/2}-1right)
&=lim_{xto-infty}-2xleft(1-frac1{8x}+oleft(frac1xright)-1right)\
&=lim_{xto-infty}-2xleft(-frac1{8x}+oleft(frac1xright)right)\
&=lim_{xto-infty}left(frac14+o(1)right)\
&=frac14\
end{align}$$

share|cite|improve this answer

answered 8 hours ago

Peter ForemanPeter Foreman

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share|cite|improve this answer

answered 8 hours ago

Peter ForemanPeter Foreman

13k11 gold badge55 silver badges2929 bronze badges

answered 8 hours ago

Peter ForemanPeter Foreman

13k11 gold badge55 silver badges2929 bronze badges

answered 8 hours ago

Peter ForemanPeter Foreman

13k11 gold badge55 silver badges2929 bronze badges

4

$begingroup$

A more elemental solution :
$$lim_{xto -infty} big( sqrt{4x^2-x} +2x big) =lim_{xto -infty} frac{-x}{sqrt{4x^2-x} -2x} =lim_{xto infty} frac{x}{sqrt{4x^2+x}+2x} =lim_{xtoinfty} frac{1}{sqrt{4+frac{1}{x}}+2} = frac{1}{sqrt{4}+2}=frac{1}{4}$$
where I used that
$$lim_{xto infty} f(x)=lim_{xto -infty} f(-x)$$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Azif00Azif00

3,03422 silver badges1414 bronze badges

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add a comment | 

4

$begingroup$

A more elemental solution :
$$lim_{xto -infty} big( sqrt{4x^2-x} +2x big) =lim_{xto -infty} frac{-x}{sqrt{4x^2-x} -2x} =lim_{xto infty} frac{x}{sqrt{4x^2+x}+2x} =lim_{xtoinfty} frac{1}{sqrt{4+frac{1}{x}}+2} = frac{1}{sqrt{4}+2}=frac{1}{4}$$
where I used that
$$lim_{xto infty} f(x)=lim_{xto -infty} f(-x)$$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Azif00Azif00

3,03422 silver badges1414 bronze badges

$endgroup$

add a comment | 

$begingroup$

A more elemental solution :
$$lim_{xto -infty} big( sqrt{4x^2-x} +2x big) =lim_{xto -infty} frac{-x}{sqrt{4x^2-x} -2x} =lim_{xto infty} frac{x}{sqrt{4x^2+x}+2x} =lim_{xtoinfty} frac{1}{sqrt{4+frac{1}{x}}+2} = frac{1}{sqrt{4}+2}=frac{1}{4}$$
where I used that
$$lim_{xto infty} f(x)=lim_{xto -infty} f(-x)$$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Azif00Azif00

3,03422 silver badges1414 bronze badges

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share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Azif00Azif00

3,03422 silver badges1414 bronze badges

answered 8 hours ago

Azif00Azif00

3,03422 silver badges1414 bronze badges

answered 8 hours ago

Azif00Azif00

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1

$begingroup$

begin{align*}
(4x^2-x)^{1/2}+2x
&=sqrt{4x^2-x}+2x
\&=frac{(sqrt{4x^2-x}+2x)(sqrt{4x^2-x}-2x)}{sqrt{4x^2-x}-2x}
\&=frac{(4x^2-x)-4x^2}{sqrt{4x^2-x}-2x}
\&=frac{4x^2-x-4x^2}{sqrt{4x^2-x}-2x}
\&=frac{-x}{sqrt{4x^2-x}-2x}
\&=frac{-x}{sqrt{(2x)^2(1-frac{1}{4x})}-2x}
\&=frac{-x}{2|x|sqrt{1-frac{1}{4x}}-2x}
\&qquad [x<0]
\&=frac{-x}{-2xsqrt{1-frac{1}{4x}}-2x}
\&=frac{1}{2sqrt{1-frac{1}{4x}}+2}
\&tofrac{1}{2sqrt{1+0}+2}
\&=frac{1}{4}.
end{align*}
(as $xto-infty$)

share|cite|improve this answer

answered 8 hours ago

mf67mf67

4555 bronze badges

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add a comment | 

1

$begingroup$

begin{align*}
(4x^2-x)^{1/2}+2x
&=sqrt{4x^2-x}+2x
\&=frac{(sqrt{4x^2-x}+2x)(sqrt{4x^2-x}-2x)}{sqrt{4x^2-x}-2x}
\&=frac{(4x^2-x)-4x^2}{sqrt{4x^2-x}-2x}
\&=frac{4x^2-x-4x^2}{sqrt{4x^2-x}-2x}
\&=frac{-x}{sqrt{4x^2-x}-2x}
\&=frac{-x}{sqrt{(2x)^2(1-frac{1}{4x})}-2x}
\&=frac{-x}{2|x|sqrt{1-frac{1}{4x}}-2x}
\&qquad [x<0]
\&=frac{-x}{-2xsqrt{1-frac{1}{4x}}-2x}
\&=frac{1}{2sqrt{1-frac{1}{4x}}+2}
\&tofrac{1}{2sqrt{1+0}+2}
\&=frac{1}{4}.
end{align*}
(as $xto-infty$)

share|cite|improve this answer

answered 8 hours ago

mf67mf67

4555 bronze badges

$endgroup$

add a comment | 

$begingroup$

begin{align*}
(4x^2-x)^{1/2}+2x
&=sqrt{4x^2-x}+2x
\&=frac{(sqrt{4x^2-x}+2x)(sqrt{4x^2-x}-2x)}{sqrt{4x^2-x}-2x}
\&=frac{(4x^2-x)-4x^2}{sqrt{4x^2-x}-2x}
\&=frac{4x^2-x-4x^2}{sqrt{4x^2-x}-2x}
\&=frac{-x}{sqrt{4x^2-x}-2x}
\&=frac{-x}{sqrt{(2x)^2(1-frac{1}{4x})}-2x}
\&=frac{-x}{2|x|sqrt{1-frac{1}{4x}}-2x}
\&qquad [x<0]
\&=frac{-x}{-2xsqrt{1-frac{1}{4x}}-2x}
\&=frac{1}{2sqrt{1-frac{1}{4x}}+2}
\&tofrac{1}{2sqrt{1+0}+2}
\&=frac{1}{4}.
end{align*}
(as $xto-infty$)

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answered 8 hours ago

mf67mf67

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$endgroup$

share|cite|improve this answer

answered 8 hours ago

mf67mf67

4555 bronze badges

answered 8 hours ago

mf67mf67

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answered 8 hours ago

mf67mf67

4555 bronze badges

1

$begingroup$

Your rationalization is almost correct. It should be

$$frac{-x}{sqrt{4x^2-x}-2x} = frac{-x}{|x|left[left(4-frac{1}{x}right)^{1/2}-frac{2x}{|x|}right]}.$$

You have assumed that $-2x = -2|x|,$ which is not true when $x$ is negative. But as $xto-infty,$ we can asume $x<0$ and thus $|x|=-x,$ so this is equal to

$$frac{1}{left(4-frac1xright)^{1/2}+2}.$$

share|cite|improve this answer

edited 7 hours ago

Mars Plastic

3,04755 silver badges3333 bronze badges

answered 8 hours ago

Thomas AndrewsThomas Andrews

134k1313 gold badges149149 silver badges303303 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Your rationalization is almost correct. It should be

$$frac{-x}{sqrt{4x^2-x}-2x} = frac{-x}{|x|left[left(4-frac{1}{x}right)^{1/2}-frac{2x}{|x|}right]}.$$

You have assumed that $-2x = -2|x|,$ which is not true when $x$ is negative. But as $xto-infty,$ we can asume $x<0$ and thus $|x|=-x,$ so this is equal to

$$frac{1}{left(4-frac1xright)^{1/2}+2}.$$

share|cite|improve this answer

edited 7 hours ago

Mars Plastic

3,04755 silver badges3333 bronze badges

answered 8 hours ago

Thomas AndrewsThomas Andrews

134k1313 gold badges149149 silver badges303303 bronze badges

$endgroup$

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$begingroup$

Your rationalization is almost correct. It should be

$$frac{-x}{sqrt{4x^2-x}-2x} = frac{-x}{|x|left[left(4-frac{1}{x}right)^{1/2}-frac{2x}{|x|}right]}.$$

You have assumed that $-2x = -2|x|,$ which is not true when $x$ is negative. But as $xto-infty,$ we can asume $x<0$ and thus $|x|=-x,$ so this is equal to

$$frac{1}{left(4-frac1xright)^{1/2}+2}.$$

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edited 7 hours ago

Mars Plastic

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answered 8 hours ago

Thomas AndrewsThomas Andrews

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edited 7 hours ago

Mars Plastic

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answered 8 hours ago

Thomas AndrewsThomas Andrews

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edited 7 hours ago

Mars Plastic

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edited 7 hours ago

Mars Plastic

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answered 8 hours ago

Thomas AndrewsThomas Andrews

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answered 8 hours ago

Thomas AndrewsThomas Andrews

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$begingroup$

For the sake of comfort, we change the sign and evaluate

$$limlimits_{xtoinfty}{(4x^2+x)^{1/2}-2x}$$

which is

$$limlimits_{xtoinfty}frac{x}{{(4x^2+x)^{1/2}+2x}}$$

or

$$limlimits_{xtoinfty}frac{1}{{(4+frac1x)^{1/2}+2}}=frac14.$$

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answered 8 hours ago

Yves DaoustYves Daoust

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$begingroup$

For the sake of comfort, we change the sign and evaluate

$$limlimits_{xtoinfty}{(4x^2+x)^{1/2}-2x}$$

which is

$$limlimits_{xtoinfty}frac{x}{{(4x^2+x)^{1/2}+2x}}$$

or

$$limlimits_{xtoinfty}frac{1}{{(4+frac1x)^{1/2}+2}}=frac14.$$

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answered 8 hours ago

Yves DaoustYves Daoust

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$begingroup$

For the sake of comfort, we change the sign and evaluate

$$limlimits_{xtoinfty}{(4x^2+x)^{1/2}-2x}$$

which is

$$limlimits_{xtoinfty}frac{x}{{(4x^2+x)^{1/2}+2x}}$$

or

$$limlimits_{xtoinfty}frac{1}{{(4+frac1x)^{1/2}+2}}=frac14.$$

share|cite|improve this answer

answered 8 hours ago

Yves DaoustYves Daoust

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answered 8 hours ago

Yves DaoustYves Daoust

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answered 8 hours ago

Yves DaoustYves Daoust

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answered 8 hours ago

Yves DaoustYves Daoust

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$begingroup$

$y:=-x$, and $ lim y rightarrow +infty$.

$(4y^2+y)^{1/2}-2y=$

$((2y+1/4)^2-1/16)^{1/2}-2y$;

$z:=2y+1/4$;

We get

$((z^2-1/16)^{1/2}-z) +1/4$;

Since

$lim_{z rightarrow infty} (z^2-1/16)^{1/2}-z)=0$ (why?), and we are done.

Note:

$(z^2-1/16)^{1/2}-(z^2)^{1/2}= $

$dfrac{-1/16}{(z^2-1/16)^{1/2}+(z^2)^{1/2}}$

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edited 7 hours ago

answered 8 hours ago

Peter SzilasPeter Szilas

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$begingroup$

$y:=-x$, and $ lim y rightarrow +infty$.

$(4y^2+y)^{1/2}-2y=$

$((2y+1/4)^2-1/16)^{1/2}-2y$;

$z:=2y+1/4$;

We get

$((z^2-1/16)^{1/2}-z) +1/4$;

Since

$lim_{z rightarrow infty} (z^2-1/16)^{1/2}-z)=0$ (why?), and we are done.

Note:

$(z^2-1/16)^{1/2}-(z^2)^{1/2}= $

$dfrac{-1/16}{(z^2-1/16)^{1/2}+(z^2)^{1/2}}$

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edited 7 hours ago

answered 8 hours ago

Peter SzilasPeter Szilas

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$begingroup$

$y:=-x$, and $ lim y rightarrow +infty$.

$(4y^2+y)^{1/2}-2y=$

$((2y+1/4)^2-1/16)^{1/2}-2y$;

$z:=2y+1/4$;

We get

$((z^2-1/16)^{1/2}-z) +1/4$;

Since

$lim_{z rightarrow infty} (z^2-1/16)^{1/2}-z)=0$ (why?), and we are done.

Note:

$(z^2-1/16)^{1/2}-(z^2)^{1/2}= $

$dfrac{-1/16}{(z^2-1/16)^{1/2}+(z^2)^{1/2}}$

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edited 7 hours ago

answered 8 hours ago

Peter SzilasPeter Szilas

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share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Peter SzilasPeter Szilas

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answered 8 hours ago

Peter SzilasPeter Szilas

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answered 8 hours ago

Peter SzilasPeter Szilas

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$begingroup$

Hint: Multiply by ${sqrt{4x^2-x}-2x}over {sqrt{4x^2-x}-2x}$

The result is ${-xover{sqrt{4x^2-x}-2x}}$

$={{-x}over {|x|sqrt{4-{1over x}}+2}}=$

${{-x}over {-xsqrt{4-{1over x}}+2}}$

${1over {sqrt{4-{1over x}}+2}}$.

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edited 3 hours ago

answered 9 hours ago

Tsemo AristideTsemo Aristide

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  $begingroup$
  $-2x$ is not under the root.
  $endgroup$
  – Tapi
  8 hours ago
  

  
  
  
  


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  $begingroup$
  $-2x=2|x|,$ at least when $x<0.$
  $endgroup$
  – Thomas Andrews
  8 hours ago
  

  
  
  
  


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  $begingroup$
  This answer is still wrong. Instead of $-frac 2x$ in the denominator, it should be simple $+2.$
  $endgroup$
  – Thomas Andrews
  6 hours ago
  

  
  
  
  

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