real analysis - Find limit $lim_{xrightarrow 0}frac{2^{sin(x)}-1}{x} = 0$

can someone provide me with some hint how to evaluate this limit?
$$\lim_{x\rightarrow 0}\frac{2^{\sin(x)}-1}{x} = 0$$
Unfortunately, I can't use l'hopital's rule
I was thinking about something like that:
$$\lim_{x\rightarrow 0}\frac{2^{\sin(x)}-1}{x} =\\\lim_{x\rightarrow 0}\frac{\ln(e^{2^{\sin(x)}})-1}{\ln(e^x)}$$ but there I don't see how to continue this way of thinking (of course if it is correct)

Answer

Hint:

For $\sin x\ne0$

$$\dfrac{2^{\sin x}-1}x=\dfrac{2^{\sin x}-1}{\sin x}\cdot\dfrac{\sin x}x$$

$$\implies\lim_{x\to0}\dfrac{2^{\sin x}-1}x=\lim_{x\to0}\dfrac{2^{\sin x}-1}{\sin x}\cdot\lim_{x\to0}\dfrac{\sin x}x$$