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

Question

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

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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?

Answer 1

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 $t=\tan x \to -1$

$$=\lim_{x \to -1}\frac{4t^2 -(1-t)(1+t^2)}{(1+t)(1+t^2)}=\lim_{x \to -1}\frac{t^3 + 3 t^2 + t - 1}{(1+t)(1+t^2)}=$$$$=\lim_{x \to -1}\frac{(t + 1) (t^2 + 2 t - 1)}{(1+t)(1+t^2)}=\lim_{x \to -1}\frac{t^2 + 2 t - 1}{1+t^2}=\frac{1-2-1}{1+1}=-1$$

Answer 2

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.

Answer 3

We try to simplify the numerator and factor out $(\sin x+\cos x)$ to avoid using L'Hospital.

$4\sin^2x \cos x-\cos x+\sin x\\ =\sin x + 3\sin^2 x\cos x+(-\cos x+\sin^2 x\cos x)\\ =\sin x + 3\sin^2 x\cos x +\cos x(\sin^2 x-1)\\ =\sin x + 3\sin^2 x\cos x -\cos^3 x+(\sin^3x-\sin^3x)\\ = \sin x (1-\sin^2x)+ 3\sin^2 x\cos x -\cos^3 x+\sin^3x\\ = \sin x \cos^2 x+ 3\sin^2 x\cos x -\cos^3 x+\sin^3x\\ = (2\sin x \cos^2 x-\sin x \cos^2 x)+ (2\sin^2 x\cos x +\sin^2 x\cos x) -\cos^3 x+\sin^3x\\ = (\sin^3x-\sin x \cos^2 x+ 2\sin^2 x\cos x) +(2\sin x \cos^2 x+\sin^2 x\cos x -\cos^3 x)\\ = \sin x(\sin^2x- \cos^2 x+ 2\sin x\cos x) +\cos x(2\sin x \cos x+\sin^2 x -\cos^2 x)\\ =(\sin x+\cos x)(\sin^2x- \cos^2 x+ 2\sin x\cos x)$

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

Answer 4

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