Calculate $\lim_{x\to \pi/4}\cot(x)^{\cot(4*x)}$ without L'Hôpital's rule How can I calculate limit
$$\lim_{x\to \pi/4}\cot(x)^{\cot(4*x)}$$
without using L'Hôpital's rule?
What I have tried so far:
I tried to use the fact that $\lim_{\alpha\to 0}(1 + \alpha)^{1/\alpha} = e$ and do the following:
$$\lim_{x\to \pi/4}\cot(x)^{\cot(4 \cdot x)} = \lim_{x\to \pi/4}(1 + (\cot(x) - 1))^{\cot(4 \cdot x)} = \lim_{x\to \pi/4}(1 + (\cot(x) - 1))^{\frac{1} {\cot(x) - 1} \cdot (\cot(x) - 1) \cdot \cot(4 \cdot x)} = \lim_{x\to \pi/4}e^{(\cot(x) - 1) \cdot \cot(4 \cdot x)} = e^{\lim_{x\to \pi/4}{(\cot(x) - 1) \cdot \cot(4 \cdot x)}} $$
But I have problems calculating limit
$$\lim_{x\to \pi/4}{(\cot(x) - 1) \cdot \cot(4 \cdot x)}$$
I tried to turn $\cot(x)$ into $\frac{\cos(x)}{\sin(x)}$ as well as turning it into $\tan(x)$, but I do not see any workaround afterwards.
I would appreciate any pieces of advice. Thank you!

We substitute $c = \cot(x)$ and we get
$$
\lim\limits_{x\to\pi/4} \cot(x)^{\cot(4x)}
=\lim\limits_{c\to 1}\; c^{\frac{c^4-6c^2+1}{4c^3-4c}}
$$
because
$$
\cot(4x) = \frac{\cot^4x-6\cot^2x+1}{4\cot^3x-4\cot x}
$$
Then we have
$$
\lim\limits_{c\to 1} \;c^{\frac{c^4-6c^2+1}{4c^3-4c}}
=\lim\limits_{c\to 1} \;\exp\left(\ln(c) \cdot \frac{c^4-6c^2+1}{4c^3-4c}\right)
=\lim\limits_{c\to 1} \;\exp\left(\frac{\ln(c)}{c-1} \cdot \frac{c^4-6c^2+1}{4c^2+4c}\right) \\
=\exp\left(\lim\limits_{c\to 1} \;\frac{\ln(c)}{c-1} \cdot \lim\limits_{c\to 1} \;\frac{c^4-6c^2+1}{4c^2+4c}\right)
=\exp\left(1\cdot\frac{-4}{8} \right)= \frac{1}{\sqrt{e}}
$$
because $\lim\limits_{c\to 1} \;\frac{\ln(c)}{c-1}$ is the derivative of $\ln(x)$ evaluated at $x=1$, which is $1$.
