Limits of $\frac{\sin^2x}{x^2}$ as $x$ approaches infinity

Question

I just want to make sure I'm on the right path with the problem. The problem is as follows:

$$\lim_{x\to\infty}\frac{\sin^2x}{x^2}$$

I rewrote it as follows:

$$\lim_{x\to\infty}\frac{(\sin x)^2}{x^2}$$

Now $\sin(x)^2$ does oscillate as $x$ approaches infinity and therefore a limit does not exist. However it oscillates between the numbers $-1$ and $1$. Since the denominator would increase without bound and the numerator would only move between $-1$ and $1$, part of me wants to say that the limit is zero.

However a smarter part of me wants to say that the limit does not exist due to the numerator. Could someone shed some light on this problem?

Answer 1

To solve the following question, recall the Squeeze Theorem:

So, as we know: $$0 \le \sin^2 x \le 1$$

If we divide by $x^2$

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

If we evaluate the limit from at either ends: $$\lim_{x \to \infty} 0 = 0$$ $$\lim_{x \to \infty} \frac 1{x^2} = 0$$

Therefore, by the squeeze theorem:

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

Answer 2

What we have is as follows: for all $x>0$, $$ 0 \leq \frac{\sin^2 x}{x^2} \leq \frac {1}{x^2} $$ Now, note that $\lim_{x \to \infty} 0 = \lim_{x \to \infty} \frac 1{x^2} = 0$. What theorem can we use here to get the answer?