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?
Best Answer
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:
$$\lim_{x \to \infty} \frac{\sin^2 x}{x^2} = 0$$