Calculate limit $\lim_{x \to 2} \frac{(2^x)-4}{\sin(\pi x)}$ without L'Hopital's rule How to calculate limit: $\lim_{x \to 2} \frac{(2^x)-4}{\sin(\pi x)}$ without L'Hopital's rule?
If $x = 2$, I get uncertainty $\frac{0}{0}$

Put $x=t+2$ and compute
$$4\lim_{t\to 0}\frac{e^{t\ln(2)}-1}{\sin(\pi t)}$$
$$=\frac{4\ln(2)}{\pi}\lim_{t\to 0}\frac{e^{t\ln(2)-1}}{t\ln(2)}\frac{\pi t}{\sin(\pi t)}$$
$$=\frac{4\ln(2)}{\pi}.$$
