Đáp án: $\lim_{x\to2}\dfrac{x^3-4\sqrt{x+2}}{x-2}=11$
Giải thích các bước giải:
$\lim_{x\to2}\dfrac{x^3-4\sqrt{x+2}}{x-2}$
$=\lim_{x\to2}\dfrac{x^3-8-4(\sqrt{x+2}-2)}{x-2}$
$=\lim_{x\to2}\dfrac{(x-2)(x^2+2x+4)-4\cdot\dfrac{x+2-2^2}{\sqrt{x+2}+2}}{x-2}$
$=\lim_{x\to2}\dfrac{(x-2)(x^2+2x+4)-4\cdot\dfrac{x-2}{\sqrt{x+2}+2}}{x-2}$
$=\lim_{x\to2}x^2+2x+4-\dfrac{4}{\sqrt{x+2}+2}$
$=2^2+2\cdot2+4-\dfrac{4}{\sqrt{2+2}+2}$
$=11$
