# How do you find the limit of sin((x-1)/(2+x^2))  as x approaches infinity?

Dec 8, 2016

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

#### Explanation:

Because $f \left(x\right) = \sin \left(x\right)$ is continuous, we have

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

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

$= \sin \left(\frac{0 - 0}{0 + 1}\right)$

$= \sin \left(0\right)$

$= 0$