Is it possible to evaluate $\lim_{x\rightarrow 0}(-1+\cos x)^{\tan x}$?

Question

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?

Answer 1

$$\cos x-1<0$$ in the neighborhood of $0$ so that the function cannot be evaluated (at best values closer and closer to $\pm1$ for rational exponents).

Hence the limit does not exist.

Answer 2

Quite long but it's correct to me.

To try with something else, you could:

1. Use Taylor Series

$$\tan(x)\approx x$$

$$\cos(x) \approx 1 - \frac{x^2}{2}$$

2. Use a different approach

For example the exponential representation

$$(\cos(x)-1)^{\tan(x)} = \text{exp}\Big( \tan(x)\log\left(-2\sin^2\frac{x}{2}\right)\Big)$$

Answer 3

You can reduce 1 step that's looking odd.

You can write simply.

$$\lim_{x\to 0} \left[-1\left(1-\cos x \right)\right]^{\tan x}$$

$$\lim_{x\to 0} \left[-1\left(2\sin^2 \frac x2\right)\right]^{\tan x}$$

$$\lim_{x\to 0} \left(-2\right)^{\tan x} . \left(\sin^2 \frac x2 \right)^{\tan x}$$