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


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?

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Do not rewrite, just look. The top wiggles, but stays small. The bottom blows up. – André Nicolas Jul 6 '14 at 0:52
Reminder that the numerator would only oscillate between $0$ and $1$ because it is $\sin^2 x$ rather than just $\sin x$. – Brad Jul 6 '14 at 2:18
The limit of $\sin^2x$ doesn't exist, but that doesn't mean the limit of $\frac{\sin^2x}{x^2}$ can't. – user2357112 Jul 6 '14 at 4:18
Related question:… – Martin Sleziak Jan 20 at 17:58
up vote 8 down vote accepted

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:

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

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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?

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