$\lim_{x\rightarrow 0} \frac{x^{\sin x} -1}{x}$ how can i compute this limit? $\lim_{x\rightarrow 0} \frac{x^{\sin x} -1}{x}$ first of all i guess that the limit at $x\rightarrow0^-$ doesn't exist (so the whole limit doesn't exist)
bur even so i would like to know how to solve the right
limit, here's how i tried
using Taylor series
$$\lim_{x\rightarrow 0} \frac{x^{\sin x} -1}{x} = \lim_{x\rightarrow 0} \frac{x^{1-\frac{x^3}{3!}+o(x^3)} }{x}=\lim_{x\rightarrow 0} \frac{x^1 x^{-\frac{x^3}{3!}}x^{o(x^3)} -1}{x}$$
unluckily I don't know how to go on because it's my first time using landau symbols.
 A: How about writing the limit as
\begin{eqnarray}
\mathcal L &=& \lim_{x\to 0^+}\frac{x^{\sin x}-1}{x} = \\
&=& \lim_{x \to 0^+}\frac{e^{\sin x \log x} -1}{x}=\\
&=&\lim_{x \to 0^+}\frac{e^{\frac{\sin x}x \cdot x\log x}-1}x=\\
&=&\lim_{x\to 0^+}\frac{e^{x\log x}-1}x=\\
&=&\lim_{x\to 0^+}\frac{e^{x\log x}-1}{x\log x}\log x\\
&=&\lim_{x \to 0^+}\log x = -\infty,
\end{eqnarray}
where we just used
$$\frac{\sin \alpha(x)}{\alpha(x)} \to 1$$
and
$$\frac{e^{\alpha(x)}-1}{\alpha(x)} \to  1\tag{1}\label{1}$$
whenever $\alpha(x) \to 0$.
As per comment below. An alternative way is to separately show that
$$\lim_{x \to 0^+} \sin x \log x = \lim_{x\to 0^+} \frac{\sin x}x x \log x = 0,$$ and then use \eqref{1}.
A: We can exploit Taylor series as follows: First, write the $x^{\sin x}$ as
$$x^{\sin x} = \exp(\log x \sin x) .$$
The Taylor series for $\sin$ at $x = 0$ is
$$\sin x \sim x + R_1(x) ,$$ where $R_1(x) \in O(x^3)$, so
$$\log x \sin x \sim x \log x + R_2(x),$$ where $R_2(x) \in O(x^3 \log x) \subset O(x^2)$. Substituting into the Taylor series $$\exp x \sim 1 + x + \frac{x^2}{2} + R_3(x) ,$$ where $R_3(x) \in O(x^3)$, gives
$$x^{\sin x} \sim 1 + x \log x + R_4(x) ,$$ where $R_4(x) \in O(x^2 \log^2 x)$. Substituting gives that
$$\frac{x^{\sin x} - 1}{x} \sim \log x + R_5(x) ,$$
where $R_5(x) \in O(x \log^2 x)$. By l'Hopital $R_5(x) \to 0$ as $x \searrow 0$, so $$\lim_{x \searrow 0} \frac{x^{\sin x} - 1}{x} = \lim_{x \searrow 0} \log x = -\infty .$$
