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)

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


For $\sin x\ne0$

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

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

  • $\begingroup$ Ok, second part is a theorem which I know, but first part? It is needed to transform it or you mean about theorem there? $\endgroup$ – VirtualUser Jan 20 at 10:53
  • $\begingroup$ Why is the limit of the remaining ratio supposed to be easier to determine than the limit of the original ratio? $\endgroup$ – Did Jan 20 at 11:11
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    $\begingroup$ 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? $\endgroup$ – Yuval Gat Jan 20 at 11:23
  • $\begingroup$ @VirtualUser, Use math.stackexchange.com/questions/942565/… $\endgroup$ – lab bhattacharjee Jan 20 at 12:45

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