Finding $\lim\limits_{x→\frac π4} \frac{\tan(4x)}{\sin(2x)-1}$

Question

I have a problem here. I try a lot of methods like make it $\dfrac{\sin2x}{2x}$ but I did not solve it. Please, if someone have just an idea about a method to solve it I will be thankful.

Answer 1

Set $\dfrac\pi4-x=y$

$$\lim_{x\to\frac\pi4}\dfrac{\tan4x}{\sin2x-1}=\lim_{y\to0}\dfrac{\tan4y}{1-\cos2y}=\lim_{y\to0}\dfrac{1+\cos2y}{\cos4y}\cdot\lim_{y\to0}\dfrac{\sin4y}{4y}\cdot\left(\dfrac{2y}{\sin2y}\right)^2\cdot\lim_{y\to0}\dfrac1 y$$

Now what can say about the existence of the following limit?

$\lim_{y\to0}\dfrac1 y$

Answer 2

Set $y=\pi/2-2x$, so $2x=\pi/2-y$ and \begin{align} \tan4x&=\tan(\pi-2y)=-\tan2y \\[4px] \sin2x&=\cos y \end{align} If $x\to\pi/4$, then $y\to0$, so your limit becomes $$ \lim_{y\to0}\frac{\tan2y}{1-\cos y}= \lim_{y\to0}\frac{2\sin y}{1-\cos y}\frac{\cos y}{\cos2y} $$ What do you know about $$ \lim_{y\to0}\frac{\sin y}{y} \qquad\text{and}\qquad \lim_{y\to0}\frac{1-\cos y}{y} $$ which are very basic limits?