# Question #d78f0

Sep 2, 2016

$= 0$

#### Explanation:

${\lim}_{x \to \infty} \frac{{\sin}^{2} x}{{x}^{2} + 1}$

We don't need to do very much here. $\sin x$ is a periodic and continuous function such that $\sin \left(x\right) \in \left[- 1 , 1\right] , x \in \left(- \infty , \infty\right)$

we can lift it out of the limit so that

${\lim}_{x \to \infty} \frac{{\sin}^{2} x}{{x}^{2} + 1}$

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

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

and the limit of the quotient is the quotient of the limits, where the limits are known, so...
$= {\lim}_{x \to \infty} \left({\sin}^{2} x\right) \frac{{\lim}_{x \to \infty} 1}{{\lim}_{x \to \infty} \left({x}^{2} + 1\right)}$

$= \left({\sin}^{2} x\right) \frac{1}{\infty} = 0$