Đáp án: $\lim_{x\to0}\dfrac{\sqrt{x^2+7}-\sqrt{4x+7}}{x^3+3x^2-x}=\dfrac{2\sqrt{7}}{7}$
Giải thích các bước giải:
$\lim_{x\to0}\dfrac{\sqrt{x^2+7}-\sqrt{4x+7}}{x^3+3x^2-x}$
$=\lim_{x\to0}\dfrac{\sqrt{x^2+7}-\sqrt{7}-(\sqrt{4x+7}-\sqrt{7})}{x^3+3x^2-x}$
$=\lim_{x\to0}\dfrac{\dfrac{x^2+7-7}{\sqrt{x^2+7}+\sqrt{7}}-\dfrac{4x+7-7}{\sqrt{4x+7}+\sqrt{7}}}{x^3+3x^2-x}$
$=\lim_{x\to0}\dfrac{\dfrac{x^2}{\sqrt{x^2+7}+\sqrt{7}}-\dfrac{4x}{\sqrt{4x+7}+\sqrt{7}}}{x^3+3x^2-x}$
$=\lim_{x\to0}\dfrac{\dfrac{x}{\sqrt{x^2+7}+\sqrt{7}}-\dfrac{4}{\sqrt{4x+7}+\sqrt{7}}}{x^2+3x-1}$
$=\dfrac{\dfrac{0}{\sqrt{0^2+7}+\sqrt{7}}-\dfrac{4}{\sqrt{4.0+7}+\sqrt{7}}}{0^2+3.0-1}$
$=\dfrac{2\sqrt{7}}{7}$
