Calculate $\lim_{x\to \pi/4}\cot(x)^{\cot(4*x)}$ without L'Hôpital's rule

$$\lim_{x\to\pi/4}\cot x^{\cot4x}=\left(\lim_{x\to\pi/4}(1+\cot x-1)^{1/(\cot x-1)}\right)^{\lim_{x\to\pi/4}{\cot4x(\cot x-1)}}$$

The inner limit converges to $e$

For the exponent, $$\lim_{x\to\pi/4}\cot4x(\cot x-1)=\lim_{x\to\pi/4}\dfrac{\cos4x}{\sin x}\cdot\lim_{x\to\pi/4}\dfrac{\cos x-\sin x}{\sin4x}$$

Now

$$\lim_{x\to\pi/4}\dfrac{\cos4x}{\sin x}=\dfrac{\cos\pi}{\sin\dfrac\pi4}=?$$

Finally

Method$\#:1$

$$F=\lim_{x\to\pi/4}\dfrac{\cos x-\sin x}{\sin4x}=\lim_{x\to\pi/4}\cos x\cdot\lim_{x\to\pi/4}\dfrac{\cot x- 1}{\sin4x}=\dfrac1{\sqrt2}\cdot\lim_{x\to\pi/4}\dfrac{\cot x- \cot\dfrac\pi4}{\sin4x-\sin\pi}$$

Method$\#:1A$

$$F=\dfrac1{\sqrt2}\cdot\dfrac{\dfrac{d(\cot x)}{dx}}{\dfrac{d(\sin4x)}{dx}}_{\text{at } x=\pi/4}$$

Method$\#:1B$

$$F=\dfrac1{\sqrt2}\cdot\lim_{x\to\pi/4}\dfrac1{\sin x\sin\dfrac\pi4} \cdot\lim_{x\to\pi/4}\dfrac{\sin\left(\dfrac\pi4-x\right)}{\sin4\left(\dfrac\pi4-x\right)}=\dfrac{\sqrt2}4$$

Method$\#:2$

set $\dfrac\pi4-x=y$ $$F=\sqrt2\lim_{x\to\pi/4}\dfrac{\sin\left(\dfrac\pi4-x\right)}{\sin4x}=\sqrt2\lim_{y\to0}\dfrac{\sin y}{\sin4\left(\dfrac\pi4-y\right)}=\dfrac{\sqrt2}4$$

I would use the ansatz $$\lim_{x\to \frac{\pi}{4}}\exp(\cot(4x)\log(\cot(x)))$$ $$=\exp\left(\lim_{x\to \frac{\pi}{4}}\frac{\cos(4x)\log(\cot(x))}{\sin(4x))}\right)$$ $$=\exp\left(\lim_{x\to \frac{\pi}{4}}\cos(4x)\lim_{x\to \frac{\pi}{4}}\frac{\log(\cot(x))}{\sin(4x)}\right)$$ $$=\exp\left(-1\lim_{x\to \frac{\pi}{4}}\frac{\log(\cot(4x))}{\sin(4x)}\right)$$

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 $1$.