# Evaluating $\lim_{x\to\pi/4} \frac{(x - \frac{\pi}{4})\sin\left(3x - \frac{3\pi}{4}\right)}{2(1-\sin2x)}$ only by using simple algebra

As I mentioned in the title, I'm not allowed to use L'hopital, series representation, and other advance method. By using trigonometric rules and simple algebra, how to evaluate this limit below:

$$\lim_{x\to\pi/4} \frac{(x - \frac{\pi}{4})\sin\left(3x - \frac{3\pi}{4}\right)}{2(1-\sin(2x))}$$

I know the answer is $$3/4$$ (I'm using L'hopital). I'm curious if that problem can be solved by using simple algebra and trigonometric rules only since that problem is for high school (they haven't learned about L'hopital). Any idea? My problem is getting rid of $$(x - \frac{\pi}{4})$$. Thanks in advance!

By using the substitution $$u=x-\frac{\pi}4$$ we get,
$$\lim_{u\to0}\frac{u\sin (3u)}{2(1-\sin(2u+\frac{\pi}2))}=\lim_{u\to0}\frac{u\sin (3u)}{2(1-\cos(2u))}=\lim_{u\to0}\frac{u\sin (3u)}{4\sin^2u}$$ Now divide numerator and denominator by $$u^2$$:
$$\frac14\lim_{u\to0}\dfrac{\dfrac{\sin (3u)}{u}}{\dfrac{\sin^2u}{u^2}}=\frac14\lim_{u\to0}\dfrac{\dfrac{\color{blue}{3}\sin (3u)}{\color{blue}{3}u}}{\dfrac{\sin^2u}{u^2}}=\frac14\lim_{u\to0}\dfrac{3.\dfrac{\sin (3u)}{3u}}{\left(\dfrac{\sin u}{u}\right)^2}$$
And by using $$\lim_{t\to0}\frac{\sin t}{t}=1$$ we have $$\frac14\times\frac{3\times1}{1^2}=\frac34$$