Evaluating $\lim_{\theta \to 0^+}\frac{\sin\theta}{\theta^2}$ 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?
 A: 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. 
A: Right, it's not necessarily valid to separate the limit like you did; but it does at least suggest the right technique: Since $$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$$
It's true that $\sin \theta / \theta > 1/2$ for all $\theta$ sufficiently small. As a result,
$$\frac{\sin \theta}{\theta^2} = \frac{\sin \theta}{\theta} \cdot \frac 1 {\theta} > \frac 1 {2 \theta}$$
Since the right side tends to infinity as $\theta \to 0^+$, the left hand side does too.
A: T. Bongers already gave the best answer, but a slightly different way to do it is to note that the limit is
$$
\lim_{\theta \to 0^+} \frac{\sin(\theta)}{\theta}\frac{1}{\theta}.
$$
You, probably, already know that 
$$
\lim_{\theta \to 0}\frac{\sin(\theta)}{\theta} = 1.
$$
So, you are multiplying something that is approaching $1$ by something that is approaching infinity. The whole thing is going to grow without bound.
