# Thread: Need help with limit

1. ## Need help with limit

Hello,

I am having trouble finding the following limit:

lim as x -> infinity of sin(x-1/x^2+2)

Factoring the denominator does not help.

Thanks so much

2. Originally Posted by tigerlily
Hello,

I am having trouble finding the following limit:

lim as x -> infinity of sin(x-1/x^2+2)

Factoring the denominator does not help.

Thanks so much
$\frac{x-1}{x^2 - 1} = \frac{1}{x+1} \to 0$ as $x \to +\infty$. And you should have a theorem about the limit of a composite functions that you can use ....

3. Originally Posted by mr fantastic
$\frac{x-1}{x^2 - 1} = \frac{1}{x+1} \to 0$ as $x \to +\infty$. And you should have a theorem about the limit of a composite functions that you can use ....
I don't see how this helps, considering $\displaystyle \frac{x-1}{x^2 - 1} \neq \frac{x - 1}{x^2 + 2}$...

Instead, $\displaystyle \lim_{x \to \infty}\frac{x - 1}{x^2 + 2} = \lim_{x \to \infty}\frac{\frac{1}{x} - \frac{1}{x^2}}{1 + \frac{2}{x^2}} = 0$.

And from the continuity of the sine function

$\displaystyle \lim_{x \to \infty}\sin{\left(\frac{x-1}{x^2 + 2}\right)} = \sin{\left[\lim_{x \to \infty}\left(\frac{x-1}{x^2 + 2}\right)\right]} = \sin{0} = 0$

4. Originally Posted by Prove It
I don't see how this helps, considering $\displaystyle \frac{x-1}{x^2 - 1} \neq \frac{x - 1}{x^2 + 2}$...

Instead, $\displaystyle \lim_{x \to \infty}\frac{x - 1}{x^2 + 2} = \lim_{x \to \infty}\frac{\frac{1}{x} - \frac{1}{x^2}}{1 + \frac{2}{x^2}} = 0$.

And from the continuity of the sine function

$\displaystyle \lim_{x \to \infty}\sin{\left(\frac{x-1}{x^2 + 2}\right)} = \sin{\left[\lim_{x \to \infty}\left(\frac{x-1}{x^2 + 2}\right)\right]} = \sin{0} = 0$
I could have sworn it was x^2 - 1 just before I posted.