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