# What is the limit of sin((x-1)/(2+x^2)) as x approaches infinity?

Apr 11, 2017

${\lim}_{x \to \infty} \sin \left(\frac{x - 1}{{x}^{2} + 2}\right) = 0$

#### Explanation:

We have that:

${\lim}_{x \to \infty} \frac{x - 1}{{x}^{2} + 2} = {\lim}_{x \to \infty} \frac{1 - \frac{1}{x}}{x + \frac{2}{x}} = 0$

As $\sin x$ is continuous in $x = 0$:

${\lim}_{x \to \infty} \sin \left(\frac{x - 1}{{x}^{2} + 2}\right) = \sin \left({\lim}_{x \to \infty} \frac{x - 1}{{x}^{2} + 2}\right) = \sin \left(0\right) = 0$