Limit of a function with sine

• Sep 9th 2009, 10:41 AM
thomasdotnet
Limit of a function with sine
Hello,

I am trying to calculate the following limit:

$\lim\limits_{x\rightarrow+\infty}{\frac{1}{x-\sqrt{x^2+1}}\sin{\frac{1}{x}}}$

So far I've multiplied by $\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2+1}}$, which yields the following:

$\lim\limits_{x\rightarrow+\infty}{\frac{x+\sqrt{x^ 2+1}}{-1}\sin{\frac{1}{x}}} = \lim\limits_{x\rightarrow+\infty}{-(x+\sqrt{x^2+1})\sin{\frac{1}{x}}}$

I don't really know how to continue from here. I'd appreciate any help.
• Sep 9th 2009, 11:23 AM
alexmahone
-delete-
• Sep 9th 2009, 12:08 PM
Opalg
Quote:

Originally Posted by thomasdotnet
I am trying to calculate the following limit:

$\lim\limits_{x\rightarrow+\infty}{\frac{1}{x-\sqrt{x^2+1}}\sin{\frac{1}{x}}}$

So far I've multiplied by $\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2+1}}$, which yields the following:

$\lim\limits_{x\rightarrow+\infty}{\frac{x+\sqrt{x^ 2+1}}{-1}\sin{\frac{1}{x}}} = \lim\limits_{x\rightarrow+\infty}{-(x+\sqrt{x^2+1})\sin{\frac{1}{x}}}$

First, $\lim_{x\to\infty}x\sin\tfrac1x = \lim_{x\to\infty}\frac{\sin\frac1x}{\frac1x} = \lim_{y\to0}\frac{\sin y}y = 1$ (where y=1/x). Then

$\lim_{x\to+\infty}-(x+\sqrt{x^2+1})\sin\tfrac1x = \lim_{x\to+\infty}-(1+\sqrt{1+x^{-2}})x\sin\tfrac1x = -2.$
• Sep 9th 2009, 02:09 PM
thomasdotnet
That solves my problem. Thank you.