<s>[INST] Is it possible to evaluate $\lim_{x\rightarrow 0}(-1+\cos x)^{\tan x}$? 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?
 [/INST] You can reduce 1 step that's looking odd.
You can write simply.
$$\lim_{x\to 0} \left[-1\left(1-\cos x \right)\right]^{\tan x}$$
$$\lim_{x\to 0} \left[-1\left(2\sin^2 \frac x2\right)\right]^{\tan x}$$
$$\lim_{x\to 0} \left(-2\right)^{\tan x} . \left(\sin^2 \frac x2 \right)^{\tan x}$$
</s>