Đáp số:
$-\infty, \sqrt{2}$
Ta tính
$\lim \dfrac{\sin(2x)}{1 + \cos^3x} = \lim \dfrac{2\sin x \cos x(1 - \cos x)}{(1 - \cos x)(1 + \cos x)(\cos^2x - \cos x + 1)}$
$= \lim \dfrac{2\sin x \cos x(1 - \cos x)}{\sin^2x(\cos^2x - \cos x + 1)}$
$= \lim \dfrac{2\cos x(1 - \cos x)}{\sin x (\cos^2x - \cos x + 1)} = \dfrac{-4}{0} = -\infty$
Lời giải:
Ta có
$\sin\left( x - \dfrac{\pi}{4} \right) = \dfrac{\sqrt{2}}{2} (\sin x - \cos x)$
Thay vào biểu thức ta có
$\underset{x \to \frac{\pi}{4}}{\lim} \dfrac{2 - 2 \cos x\sqrt{2}}{\sin x - \cos x} = \underset{x \to \frac{\pi}{4}}{\lim} \dfrac{(2-2\cos x \sqrt{2})(2 + 2\cos x \sqrt{2})}{(\sin x - \cos x)(2 + 2\cos x \sqrt{2})}$
$= \underset{x \to \frac{\pi}{4}}{\lim} \dfrac{4 - 8 \cos^2x}{2(\sin x - \cos x)(1 + \cos x \sqrt{2})}$
$= \underset{x \to \frac{\pi}{4}}{\lim} \dfrac{-2(2\cos^2x-1)}{(\sin x - \cos x)(1 + \cos x \sqrt{2})}$
$= \underset{x \to \frac{\pi}{4}}{\lim} \dfrac{-2\cos(2x)}{(\sin x - \cos x)(1 + \cos x \sqrt{2})}$
$= \underset{x \to \frac{\pi}{4}}{\lim} \dfrac{-2(\cos^2x - \sin^2x)}{(\sin x - \cos x)(1 + \cos x \sqrt{2})}$
$= \underset{x \to \frac{\pi}{4}}{\lim} \dfrac{2 (\cos x + \sin x)}{1 + \cos x \sqrt{2}}$
Thay $x = \dfrac{\pi}{4}$ vào ta có
$\underset{x \to \frac{\pi}{4}}{\lim} \dfrac{2 - 2 \cos x\sqrt{2}}{\sin x - \cos x} = \dfrac{2 . \sqrt{2}}{1 + 1} = \sqrt{2}$
