# Evaluating the limit of some undefined trigonometric function using Squeeze theorem

I want to know how to do a problem like this. What steps are taken and why.

I have some function:

$$\lim_{h\to3} \ \sin{\left(\frac{1}{h-3}\right)} \ e^h \ (h-3)^2$$

• If you knew the following inequality $\frac{\sin(x)}{x} \leq 1$, could you tackle this problem? – Demophilus Jun 8 '17 at 17:29
• Try drawing a graph of @rt6's solution to see what is really going on here. – John Joy Jun 10 '17 at 1:20

We know that $-1\leq \sin(x)\leq 1$ for every $x\in\mathbb{R}$.
It follows then for every value of $h\in \mathbb{R}$ except at $h=3$,
$$-e^h(h-3)^2\leq \sin\left(\frac{1}{h-3}\right)e^h(h-3)^2\leq e^h(h-3)^2$$
We can now apply the squeeze theorem as $h$ tends to $3$ to obtain that the desired limit is zero.