Answer
$$\lim_{x\to2}\frac{\sin(x^2-4)}{x-2}=4$$
Work Step by Step
$$A=\lim_{x\to2}\frac{\sin(x^2-4)}{x-2}$$
$$A=\lim_{x\to2}\frac{\sin(x^2-4)}{x^2-4}\times\lim_{x\to2}\frac{x^2-4}{x-2}$$
For $x\ne\pm2$, $x^2-4\ne0$ and $\lim_{x\to2}(x^2-4)=0$. So we have $\lim_{x\to2}\frac{\sin(x^2-4)}{x^2-4}=1$.
$$A=1\times\lim_{x\to2}\frac{(x-2)(x+2)}{x-2}$$
$$A=\lim_{x\to2}(x+2)$$
$$A=2+2=4$$
