# Find limit $\lim_{x\rightarrow 0}\frac{2^{\sin(x)}-1}{x} = 0$

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)

• The limit is precisely the definition of $f'(0)$ where $f(x) = 2^{\sin(x)}$. This technically doesn't use L'Hopital's rule! ;-) – Theo Bendit Jan 20 '19 at 10:58
• "Unfortunately, I can't use l'hopital's rule" Correction: Fortunately, you cannot use LH. – Did Jan 20 '19 at 11:10
• Just combine $\lim_{z\to 0}\frac{a^z-1}{z}=\log a$ with $\lim_{w\to 0}\frac{\sin w}{w}=1$ to get $\color{red}{\log 2}\neq 0$. – Jack D'Aurizio Jan 20 '19 at 18:44

For $$\sin x\ne0$$
$$\dfrac{2^{\sin x}-1}x=\dfrac{2^{\sin x}-1}{\sin x}\cdot\dfrac{\sin x}x$$
$$\implies\lim_{x\to0}\dfrac{2^{\sin x}-1}x=\lim_{x\to0}\dfrac{2^{\sin x}-1}{\sin x}\cdot\lim_{x\to0}\dfrac{\sin x}x$$
• The former limit $\lim\limits_{x\to 0}\frac{2^{\sin x}-1}{\sin x}$ can be solved by substituting $t=\sin x$. You then get $\lim\limits_{t\to 0}\frac{2^t-1}{t}$. Can you proceed? – Yuval Gat Jan 20 '19 at 11:23