# What is the limit of sin^4(x)/x^0.5 as x goes to infinity?

Oct 16, 2015

It is $0$

#### Explanation:

$0 \le {\sin}^{4} x \le 1$ for all $x$

${x}^{0.5} > 0$ for $x > 0$, so we can divide in the inequality without changing the directions of the inequalities.

$0 \le {\sin}^{4} \frac{x}{x} ^ 0.5 \le \frac{1}{x} ^ 0.5$ for all $x$

We note that ${\lim}_{x \rightarrow \infty} 0 = {\lim}_{x \rightarrow \infty} \frac{1}{x} ^ 0.5 = 0$,

so, by the squeeze theorem, ${\lim}_{x \rightarrow \infty} {\sin}^{4} \frac{x}{x} ^ 0.5 = 0$