Calculate limit $\lim_{x \to 2} \frac{(2^x)-4}{\sin(\pi x)}$ without L'Hopital's rule

Question

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 $\frac{0}{0}$

Answer 1

$\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} $

Answer 2

HINT:

$$\frac{2^x-4}{\sin(\pi x)}=\left(\frac{2^x-4}{x-2}\right)\,\left(\frac{x-2}{\sin(\pi (x-2))}\right) \tag 1$$

The limit of the first parenthetical term in $(1)$ is the derivative of $2^x$ at $x=2$. And $\lim_{\theta \to 0}\frac{\sin(\theta)}{\theta}=1$.

Answer 3

Put $x=t+2$ and compute

$$4\lim_{t\to 0}\frac{e^{t\ln(2)}-1}{\sin(\pi t)}$$

$$=\frac{4\ln(2)}{\pi}\lim_{t\to 0}\frac{e^{t\ln(2)-1}}{t\ln(2)}\frac{\pi t}{\sin(\pi t)}$$

$$=\frac{4\ln(2)}{\pi}.$$