TITLE: Evaluating $\lim_{\theta \to 0^+}\frac{\sin\theta}{\theta^2}$
QUESTION [6 upvotes]: How do I evaluate $$\lim_{\theta \to 0^+}\frac{\sin\theta}{\theta^2}?$$
I tried the following:
$$\lim_{\theta \to 0^+}\frac{\sin\theta}{\theta^2} = \lim_{\theta \to 0^+}\frac{1}{\theta}\cdot \lim_{\theta \to 0^+}\frac{\sin\theta}{\theta} = \lim_{\theta \to 0^+}\frac{1}{\theta} = +\infty$$
However, I feel that there is an error with my work, since I believe it isn't acceptable to separate a limit when it separates into something that has a value of infinity. Is there an issue with my work here?

REPLY [4 votes]: You may try in such a way: $\forall \epsilon >0$ we have $\lim_{\theta\rightarrow 0^{+}} \frac{sin(\theta)}{\theta^2} \ge \lim_{\theta\rightarrow 0^{+}} \frac{sin(\theta)}{\theta\epsilon} = \frac{1}{\epsilon}$. since we know limit $\ge \frac{1}{\epsilon}$ for all $\epsilon$, we can conclude that  $\lim_{\theta\rightarrow 0^{+}} \frac{sin(\theta)}{\theta^2} = \infty$ or simply say the limit does not exist.