Find $ \lim_{x \to 2} (x^2-4)/\sin(x-2)$ without using l'hopitals rule

Question

This is an indeterminate form and I think I should use the fact that $x-2/\sin(x-2) = 1 \;$but idk how to do that.

Answer 1

Hint:

While $\alpha\sim 0$, for example when $\alpha\to 0$, then $\sin(\alpha)\sim \alpha$. Here $x\to 2$ so we have $$(x-2)\to 0$$

Answer 2

$$ \lim_{x \to 2} \frac{x^2-4}{\sin(x-2)} = \lim_{x\to 2} \left( (x+2) \frac{x-2}{\sin(x-2)} \right) = \left( \lim_{x\to2} (x+2) \right) \left( \lim_{x\to2} \frac{x-2}{\sin(x-2)} \right) $$

The second limit may be written as $\displaystyle\lim_{u\to0} \frac u {\sin u}$ and you've probably seen that limit evaluated by squeezing.