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 $\frac{0}{0}$

$\lim_{x \to 2} \frac{(2^x)-4}{\sin(\pi x)}
$
Putting $x = y+2$,
$\begin{array}\\
\dfrac{(2^x)-4}{\sin(\pi x)}
&=\dfrac{(2^{y+2})-4}{\sin(\pi (y+2))}\\
&=4\dfrac{(2^{y})-1}{\sin(\pi y)}
\qquad\text{since } 2^{y+2} = 4\cdot 2^y
 \text{ and }\sin(\pi (y+2))=\sin(\pi y + 2\pi)=\sin(\pi y)\\
&=4\dfrac{e^{y\ln 2}-1}{\sin(\pi y)}\\
&\approx 4\dfrac{y\ln 2}{\pi y}
\qquad\text{since } e^z \approx 1+z \text{ and } \sin(z) \approx z
\text{ for small }z\\
&= \dfrac{4\ln 2}{\pi }
\end{array}
$
