Đáp án: $\lim_{x\to0}\dfrac{\sqrt{2x+1}-\sqrt[3]{x^2+1}}{\sin x}=1$
Giải thích các bước giải:
$\lim_{x\to0}\dfrac{\sqrt{2x+1}-\sqrt[3]{x^2+1}}{\sin x}$
$=\lim_{x\to0}\dfrac{\sqrt{2x+1}-1-(\sqrt[3]{x^2+1}-1)}{\sin x}$
$=\lim_{x\to0}\dfrac{\dfrac{2x+1-1}{\sqrt{2x+1}+1}-\dfrac{x^2+1-1}{\sqrt[3]{x^2+1}^2+\sqrt[3]{x^2+1}+1}}{\sin x}$
$=\lim_{x\to0}\dfrac{\dfrac{2x}{\sqrt{2x+1}+1}-\dfrac{x^2}{\sqrt[3]{x^2+1}^2+\sqrt[3]{x^2+1}+1}}{\sin x}$
$=\lim_{x\to0}\dfrac{\dfrac{2}{\sqrt{2x+1}+1}-\dfrac{x}{\sqrt[3]{x^2+1}^2+\sqrt[3]{x^2+1}+1}}{\dfrac{\sin x}{x}}$
$=\dfrac{\dfrac{2}{\sqrt{2.0+1}+1}-\dfrac{0}{\sqrt[3]{0^2+1}^2+\sqrt[3]{0^2+1}+1}}{1}$
$=1$
