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

Mar 21, 2018

${\lim}_{x \to \infty} {\sin}^{2} \frac{x}{x} ^ 2 = 0$

#### Explanation:

For limits involving trigonometric functions divided by something, we're best off using the Squeeze Theorem, which tells us that if we have a function $f \left(x\right) ,$ we can definite $h \left(x\right)$ and $g \left(x\right)$ such that

$h \left(x\right) \le f \left(x\right) \le g \left(x\right)$

And that if

${\lim}_{x \to a} h \left(x\right) = {\lim}_{x \to a} g \left(x\right) ,$ then ${\lim}_{x \to a} f \left(x\right) = {\lim}_{x \to a} g \left(x\right) = {\lim}_{x \to a} g \left(x\right) .$

In other words, if the two functions between which $f \left(x\right)$ lies have the same limit when approaching the same value, $f \left(x\right)$ must also have that limit.

So, recall that

$- 1 \le \sin x \le 1$

This means that

$0 \le {\sin}^{2} x \le 1$ (Sine squared can only be positive/zero, and less than one).

We have ${\sin}^{2} \frac{x}{x} ^ 2 ,$ so to accommodate this in our inequality, we can divide everything by ${x}^{2} :$

$\frac{0}{x} ^ 2 \le {\sin}^{2} \frac{x}{x} ^ 2 \le \frac{1}{x} ^ 2$

$0 \le {\sin}^{2} \frac{x}{x} ^ 2 \le \frac{1}{x} ^ 2$

Now, let's take the limits as $x \to \infty$ of $h \left(x\right) = 0 , g \left(x\right) = \frac{1}{x} ^ 2$:

${\lim}_{x \to \infty} 0 = 0$

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

So, since the limits of $h \left(x\right) , g \left(x\right)$ are equal, the limit of $f \left(x\right) = {\sin}^{2} \frac{x}{x} ^ 2$ must also be the limits of $h \left(x\right) , g \left(x\right)$.

Thus,

${\lim}_{x \to \infty} {\sin}^{2} \frac{x}{x} ^ 2 = 0$