Limit of the form 0 $\cdot$ inf

Question

I have been trying for hours to solve the following limit without using De L'Hospital theorem, but I got stuck. Can you help me? Thanks in advance! (The solution should be $-\sqrt2$)

$\lim_{x\rightarrow\pi/4}(2\sin(x)-\sqrt2)\tan(x-3\pi/4)$

Answer 1

$$2\sin x-\sqrt2=2\left(\sin x-\sin\dfrac\pi4\right)$$

Use https://mathworld.wolfram.com/ProsthaphaeresisFormulas.html

$$\tan(x-3\pi/4)=\tan(x+\pi/4-\pi)=\cdots=\dfrac{\sin(x+\pi/4)}{\cos(x+\pi/4)}$$

Now $\cos(x+\pi/4)=\sin(\pi/4-x)=-2\sin(x-\pi/4)$

Finally use $\sin2y=2\sin y\cos y$

As $x\to\pi/4,x\ne\pi/4,$

$\sin\dfrac{4x-\pi}8\ne0$ so can be cancelled out safely