Đáp án:
\(\lim\limits_{x\to \tfrac{\pi}{3}}\dfrac{\tan^3x - 3\tan x}{\cos\left(x + \dfrac{\pi}{6}\right)}=-24\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad \lim\limits_{x\to \tfrac{\pi}{3}}\dfrac{\tan^3x - 3\tan x}{\cos\left(x + \dfrac{\pi}{6}\right)}\\
= \lim\limits_{x\to \tfrac{\pi}{3}}\dfrac{\sin^3x - 3\cos^2x.\sin x}{\cos^3x\cos\left(x + \dfrac{\pi}{6}\right)}\\
= \lim\limits_{x\to \tfrac{\pi}{3}}\dfrac{1}{\cos^3x}\cdot \lim\limits_{x\to \tfrac{\pi}{3}}\dfrac{\sin^3x - 3\cos^2x.\sin x}{\cos\left(x + \dfrac{\pi}{6}\right)}\\
= \dfrac{1}{\cos^3\dfrac{\pi}{3}}\cdot \lim\limits_{x\to \tfrac{\pi}{3}}\dfrac{3\cos^3x - 9\cos x.\sin^2x}{\sin\left(x + \dfrac{\pi}{6}\right)}\qquad (l'Hôpital)\\
= 8\cdot \dfrac{3\cos^3\dfrac{\pi}{3} - 9\cos\dfrac{\pi}{3}.\sin^2\dfrac{\pi}{3}}{\sin\left(\dfrac{\pi}{3} + \dfrac{\pi}{6}\right)}\\
= 8\cdot (-3)\\
= -24
\end{array}\)
