Evaluate $\lim_{x \to 3\pi/4}\frac{4\sin^2x \cos x-\cos x+\sin x}{\sin x+\cos x}$ 
Evaluate $$\lim_{x \to 3\pi/4}\frac{4\sin^2x \cos x-\cos x+\sin x}{\sin x+\cos x}$$


Of course L-hospitals rule kills it.The question is how to do without L-hospitals rule.The substituition $x-3\pi/4=h$ seems promising but only complicates it further.Traditional methods of multiplying by congugate don't apply here.
How do i find the limit then?

We try to simplify the numerator and factor out $(\sin x+\cos x)$ to avoid using L'Hospital.
$4\sin^2x \cos x-\cos x+\sin x\\
=\sin x + 3\sin^2 x\cos x+(-\cos x+\sin^2 x\cos x)\\
=\sin x + 3\sin^2 x\cos x +\cos x(\sin^2 x-1)\\
=\sin x + 3\sin^2 x\cos x -\cos^3 x+(\sin^3x-\sin^3x)\\
= \sin x (1-\sin^2x)+ 3\sin^2 x\cos x -\cos^3 x+\sin^3x\\
= \sin x \cos^2 x+ 3\sin^2 x\cos x -\cos^3 x+\sin^3x\\
= (2\sin x \cos^2 x-\sin x \cos^2 x)+ (2\sin^2 x\cos x +\sin^2 x\cos x) -\cos^3 x+\sin^3x\\
= (\sin^3x-\sin x \cos^2 x+ 2\sin^2 x\cos x) +(2\sin x \cos^2 x+\sin^2 x\cos x -\cos^3 x)\\
= \sin x(\sin^2x- \cos^2 x+ 2\sin x\cos x) +\cos x(2\sin x \cos x+\sin^2 x -\cos^2 x)\\
=(\sin x+\cos x)(\sin^2x- \cos^2 x+ 2\sin x\cos x)$

$\therefore \lim_{x \to 3\pi/4}\frac{4\sin^2x \cos x-\cos x+\sin x}{\sin x+\cos x}=\lim_{x \to 3\pi/4}(\sin^2x- \cos^2 x+ 2\sin x\cos x)=-1$
