Find: $\displaystyle \lim_{x\to \frac{\pi}{4}} \frac{\sin 4x}{1-\sqrt{2} \cos x}$ (no L'Hopital)
By L'Hopital it is easy to show that this limit is $-4$. But I'm not finding the right approach to prove without using it.
Hints and answers appreciated. Sorry if this is a duplicate.
On
\begin{align}\lim_{x\to\frac\pi4}\frac{\sin(4x)}{1-\sqrt2\cos(x)}&=\lim_{x\to\frac\pi4}\frac{2\sin(2x)\cos(2x)}{1-\sqrt2\cos(x)}\\&=\lim_{x\to\frac\pi4}\frac{4\sin(x)\cos(x)\bigl(\cos^2(x)-\sin^2(x)\bigr)}{1-\sqrt2\cos(x)}\\&=\lim_{x\to\frac\pi4}\frac{4\sin(x)\cos(x)\bigl(2\cos^2(x)-1\bigr)}{1-\sqrt2\cos(x)}\\&=-\lim_{x\to\frac\pi4}4\sin(x)\cos(x)\bigl(1+\sqrt2\cos(x)\bigr)\\&=-4.\end{align}
On
Transform your expression like this:
$$\begin{align}\frac{\sin 4x}{1-\sqrt{2} \cos x}&=\frac{2\sin 2x\cos 2x(1+\sqrt{2}\cos x)}{1-2\cos^2x}\\&=\frac{2\sin 2x\cos 2x(1+\sqrt{2}\cos x)}{-\cos 2x}\\&=-2\sin 2x(1+\sqrt{2}\cos x)\end{align}$$
then let $x\to\frac{\pi}{4}$.
On
A simple way is to use $x=t+\pi/4$ that brings the limit in the form $$ \lim_{t\to0}\frac{\sin(4t+\pi)}{1-\cos t+\sin t}= \lim_{t\to0}\frac{-4\sin t\cos t\cos2t}{1-\cos t+\sin t} $$ and use that $$ \lim_{t\to0}\frac{1-\cos t+\sin t}{\sin t}= 1+\lim_{t\to 0}\frac{1-\cos t}{t}\frac{t}{\sin t}=1 $$
On
$$F=\cos A\cdot\lim_{x\to A}\dfrac{\sin4X-\sin4A}{\cos A-\cos x}$$
Method$\#1:$
$$F=-\cos A\cdot\dfrac{\lim_{x\to A}\dfrac{\sin4x-\sin4A}{x-A}}{\lim_{x\to A}\dfrac{\cos x-\cos A}{x-A}}-\cos A\cdot\dfrac{\dfrac{d(\sin4x)}{dx}_{(\text{ at } x=A)}}{\dfrac{d(\cos x)}{dx}_{(\text{ at } x=A)}}$$
Method$\#2:$ $$F=\cos A\cdot\lim_{x\to A}\dfrac{2\sin2(X-A)\cos2(x+A)}{2\sin\dfrac{x-A}2\sin\dfrac{x+A}2}$$
Now use $\sin2(X-A)=2\sin(x-A)\cos(x-A)=4\sin\dfrac{x-A}2\cos\dfrac{x-A}2\cos(x-A)$
Use $$\sin4x=2\sin2x\cos2x=2\sin2x(2\cos^2x-1)=2\sin2x(\sqrt2\cos{x}-1)(\sqrt2\cos{x}+1).$$ I got $-4$.