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

$\therefore \lim_{x \to 3\pi/4}\frac{4\sin^2x \cos x-\cos x+\sin x}{\sin x+\cos x}=\lim_{x \to 3\pi/4}(\sin^2x- \cos^2 x+ 2\sin x\cos x)=-1$
A: If you make $x=t+\frac 34 \pi$, the numerator becomes
$$\cos \left(t+\frac{\pi }{4}\right)-\cos \left(3 t+\frac{\pi }{4}\right)$$ and the denominator
$$-\sqrt{2} \sin (t)$$ Expanding as series around $t=0$ will give
$$\frac{\sqrt{2} t+2 \sqrt{2} t^2+O\left(t^3\right) } {-\sqrt{2} t+O\left(t^3\right) }=-1-2 t+O\left(t^2\right)$$ which shows the limit and how it is approached.
A: We have that
$$\lim_{x \to 3\pi/4}\frac{4\sin^2x \cos x-\cos x+\sin x}{\sin x+\cos x}= \lim_{x \to 3\pi/4}\frac{4\left(1-\frac 1{1+\tan^2 x}\right) -1+\tan x}{\tan x+1}=$$
$$=\lim_{x \to 3\pi/4}\frac{4\tan^2 x -(1-\tan x)(1+\tan^2 x)}{(1+\tan x)(1+\tan^2x)}=$$
and by $t=\tan x \to -1$
$$=\lim_{x \to -1}\frac{4t^2 -(1-t)(1+t^2)}{(1+t)(1+t^2)}=\lim_{x \to -1}\frac{t^3 + 3 t^2 + t - 1}{(1+t)(1+t^2)}=$$$$=\lim_{x \to -1}\frac{(t + 1) (t^2 + 2 t - 1)}{(1+t)(1+t^2)}=\lim_{x \to -1}\frac{t^2 + 2 t - 1}{1+t^2}=\frac{1-2-1}{1+1}=-1$$
A: $$L=\lim_{x \to 3\pi/4}\frac{4\sin^2x \cos x-\cos x+\sin x}{\sin x+\cos x}$$
$$L= \lim_{x \to 3\pi/4}4\sin^2 (x)+\frac{-4\sin^3 x -2\cos x}{\sin x+\cos x}+1$$
$$L=3+ \lim_{x \to 3\pi/4}\frac{-4\sin^3 x -2\cos x}{\sin x+\cos x}$$
Note that:
$$-4\sin ^3 x=\sin(3x)-3\sin x$$
$$L=1+ \lim_{x \to 3\pi/4}\frac{\sin(3 x) -\sin x}{\sin x+\cos x}$$
$$L=1+ \lim_{x \to 3\pi/4}\frac{2\sin( x) \cos( 2x)}{\sin x+\cos x}$$
It's easy to finish now. You can simplify since:
$$\cos (2x)=\cos^2 x- \sin^2 x$$
$$L=1+ 2\lim_{x \to 3\pi/4}\sin( x) (\cos x -\sin x)$$
$$\implies L=-1$$
