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limit of ((x(sqrt(x+2)))/sqrt(x+1))-x is 1/2? Why?


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#1 babipsylon

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Posted 28 July 2016 - 08:17 PM

I need this limit ((x(sqrt(x+2)))/sqrt(x+1))-x to calculate the asymptote of this function: ((x(sqrt(x+2)))/sqrt(x+1)).
which, according to the class notes: y=x+1/2
with a = 1, b = 1/2
However, the online math calculators say that currently no steps are supported to show for this kind of problem.
 
I calculated this limit as x*sqrt(1)-x = 1, but apparently the correct answer is 1/2.
 
What is my mistake?
 
Thx in advance.
 
r

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#2 mathematic

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Posted 28 July 2016 - 09:30 PM

\frac{x(\sqrt{x+2})}{\sqrt{x+1}}-x=\frac{x}{\sqrt{x+1}}(\sqrt{x+2}-\sqrt{x+1})=
\frac{x}{\sqrt{x+1}(\sqrt{x+2}+\sqrt{x+1})}(x+2-(x+1)) 

Limit = 1/2, since the denominator ->2x.


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#3 renerpho

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Posted 28 September 2016 - 03:49 AM

You should use l'Hôpital's rule for a rigorous proof:
\lim_{x\to\infty} {\frac{x\sqrt{x+2}}{\sqrt{x+1}}-x}=\lim_{x\to\infty} {\frac{x}{\sqrt{(x+1)(x+2)}+x+1}} (see the previous post).

Using l'Hôpital's rule, this equals \lim_{x\to\infty} {\frac{1}{(\sqrt{(x+1)(x+2)})'+1}}=\lim_{x\to\infty} {\frac{1}{\frac{2x+3}{2\sqrt{(x+1)(x+2)}}+1}}. Assuming the limit exists, this is equal to \frac{1}{\lim_{x\to\infty} {\frac{2x+3}{2\sqrt{(x+1)(x+2)}}}+1}.

You then evaluate \lim_{x\to\infty} {\frac{2x+3}{2\sqrt{(x+1)(x+2)}}}: Using l'Hôpital again, this is equal to \lim_{x\to\infty} {\frac{2}{\frac{2(2x+3)}{2\sqrt{(x+1)(x+2)}}}}=\lim_{x\to\infty} {\frac{2\sqrt{(x+1)(x+2)}}{2x+3}}. It follows that this limit is 1, and therefore \lim_{x\to\infty} {\frac{x\sqrt{x+2}}{\sqrt{x+1}}-x}=\frac{1}{1+1}=\frac{1}{2}.


Edited by renerpho, 28 September 2016 - 03:52 AM.

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#4 HallsofIvy

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Posted 31 December 2016 - 02:25 AM

  Why use L'Hopital's rule?  Mathematic's answer is much simpler.


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#5 ecoli

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Posted 31 December 2016 - 03:58 PM

You can use wolframalpha to visualize the function, fyi:

 

https://www.wolframa...))/sqrt(x+1))-x


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