How to calculate $\lim \limits_{x \to 0^{+}} (\sin x)^{e^{x}-1} $ with Taylor series? I want to calculate $\lim \limits_{n \to 0^{+}} (\sin x)^{e^{x}-1} $ by using Taylor's Series, and here is what I did so far, and correct me if I'm wrong:


*

*$\sin x = x + o(x)$

*$e^{x}-1= x + o(x)$


then I substituted these two expressions in the initial limit and got something like this:
$\lim \limits_{x \to 0^{+}} (x+o(x))^{x+o(x)}$
but how do I continue from here? 
 A: $$\lim_{x \to 0^{+}} (\sin x)^{e^{x}-1}=\lim_{x\to0^+}e^{\frac{e^x-1}x.x\ln(\sin x)}$$
Now:
$$\lim_{x\to0}\frac{e^x-1}x=1\\\lim_{x\to0^+}x\ln(\sin x)=0$$
the last one because:
$$\lim_{x\to0^+}x\ln(\sin x)=\lim_{x\to0^+}\frac{\ln(\sin x)}{1/x}=\lim_{x\to0^+}\frac{\cot x}{-1/x^2}=\lim_{x\to0^+}-x.\frac{x}{\tan x}=0$$
as:
$$\lim_{x\to0}\frac{\tan x}x=1$$
So limit is $1$.
A: The first order Taylor expansions $\sin(u) \underset{u\to0}= u + o(u)$ and $\ln(1+u) \underset{u\to 0}= u + o(u)$ give for $x \to 0^+$,
$$
\log(\sin(x)) = \log(x+o(x)) = \log x + \log(1+o(1)) = \log x + o(1).
$$
Therefore, using $e^u \underset{x\to 0} = 1 + x +o(x)$, we obtain
\begin{align}
(\sin(x))^{e^x - 1} &= \exp[(e^x - 1)\log (\sin(x))]\\
& = \exp[(x+o(x))(\log x +o(1))]\\
& = \exp[x\log x + o(x\log x)]\\
& = \exp[o(1)]\\
& = 1 + o(1)
\end{align}
since $x\log x = o(1)$.
A: If $f(x) = (\sin x)^{e^{x} - 1}$ then $f(x)$ is defined in some interval of type $(0, a)$ and hence it makes sense to ask for its limit as $x \to 0^{+}$. If $L$ is the desired limit then
\begin{align}
\log L &= \log\left(\lim_{x \to 0^{+}}(\sin x)^{e^{x} - 1}\right)\notag\\
&= \lim_{x \to 0^{+}}\log(\sin x)^{e^{x} - 1}\text{ (by continuity of log)}\notag\\
&= \lim_{x \to 0^{+}}(e^{x} - 1)\log \sin x\notag\\
&= \lim_{x \to 0^{+}}\frac{e^{x} - 1}{x}\cdot x\cdot\log \left(x\cdot\frac{\sin x}{x}\right)\notag\\
&= \lim_{x \to 0^{+}}1\cdot x\cdot\left\{\log x + \log\left(\frac{\sin x}{x}\right)\right\}\notag\\
&= \lim_{x \to 0^{+}}x\log x + x\log\left(\frac{\sin x}{x}\right)\notag\\
&= \lim_{x \to 0^{+}}x\log x + 0\cdot\log 1\notag\\
&= \lim_{x \to 0^{+}}x\log x\notag\\
&= 0\notag
\end{align}
Hence $L = e^{0} = 1$. Here we have used the standard limits $$\lim_{x \to 0}\frac{\sin x}{x} = 1,\,\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1,\, \lim_{x \to 0^{+}}x \log x = 0$$
