Let's say we have the following limit:
$$\lim\_{x\rightarrow 0}(-1+\cos x)^{\tan x}$$
Would the following solution be correct?

> 
> The solution is **incorrect**, please see the correction of @YvesDaoust
> 
> 
> 

\begin{align}
\lim\_{x \rightarrow 0}(-1+\cos x)^{\tan x}
&= \lim\_{x \rightarrow 0}\left((1-\cos x)^{\tan x}\cdot (-1)^{\tan x}\right) \\
&= \lim\_{x \rightarrow 0}\left(1-\left(1-2\sin^2\left(\frac{x}{2}\right)\right)^{\tan x}\right) \cdot \lim\_{x \rightarrow 0}(-1)^{\tan x} \\
&= \lim\_{x \rightarrow 0}\left(2\sin^2 \left(\frac{x}{2}\right)\right)^{\tan x} \cdot 1\\
&= \lim\_{x \rightarrow 0}2^{\tan x} \cdot \lim\_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{2\tan x} \\
&= 1 \cdot \lim\_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{2\tan x} \\
&= \lim\_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{2\frac{\sin x}{\cos x}} \\
&= \lim\_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{\frac{4\sin (\frac{x}{2})\cdot \cos (\frac{x}{2})}{\cos x}} \\
&= \lim\_{x \rightarrow 0}\left(\sin \left(\frac{x}{2}\right)^{\sin \left(\frac{x}{2}\right)}\right)^{\frac{4\cos\frac{x}{2}}{\cos x}} \\
&= \left(\lim\_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{\sin \left(\frac{x}{2}\right)}\right)^{\lim\_{x \rightarrow 0}\frac{4\cos\frac{x}{2}}{\cos x}} \\
&= \left(\lim\_{u \rightarrow 0}u^u\right)^{4} \\
&= 1^4 \\
&= 1 \\
\end{align}

The result seems to be correct, but the way leading to it seems to be quite lengthy. Am I doing something redundant?