Giải thích các bước giải:
Ta có :
$L=\lim_{x\to \pi}\dfrac{\sin 2x}{1+\cos x}$
$=\lim_{x\to \pi}\dfrac{2\sin x\cos x}{1+\cos x}$
$=\lim_{x\to \pi}\dfrac{2\sin x(\cos x+1)-2\sin x}{1+\cos x}$
$=\lim_{x\to \pi}2\sin x-\dfrac{2\sin x}{1+\cos x}$
$=\lim_{x\to \pi}2\sin x-\dfrac{2\sqrt{1-\cos x^2}}{1+\cos x}$
$=\lim_{x\to \pi}2\sin x-\dfrac{2\sqrt{(1+\cos x)(1-\cos x)}}{1+\cos x}$
$=\lim_{x\to \pi}2\sin x-\dfrac{2\sqrt{1-\cos x}}{\sqrt{1+\cos x}}$
$=2\sin\pi-\dfrac{2\sqrt{1-\cos\pi}}{\sqrt{1+\cos\pi}}$
$=0-\dfrac{2\sqrt{2}}{\sqrt{1-1}}$
$=-\infty$
$\to\lim_{x\to \pi}\dfrac{\sin 2x}{1+\cos^3x}$
$=\lim_{x\to \pi}\dfrac{\sin 2x}{1+\cos x}.\dfrac{1}{1-\cos x+\cos^2x}$
$=-\infty.\dfrac{1}{1-(-1)+(-1)^2}$
$=-\infty.\dfrac{1}{3}$
$=-\infty$
