Ta có:
$\mathop{\lim}\limits_{x\to 0}\dfrac{\sqrt{2x+1}-\sqrt[3]{x^2+1}}{\sin x}$
$=\mathop{\lim}\limits_{x\to 0}\dfrac{(\sqrt{2x+1}-1)-(\sqrt[3]{x^2+1}-1)}{\sin x}$
$=\mathop{\lim}\limits_{x\to 0}\dfrac{\dfrac{2x}{\sqrt{2x+1}+1}-\dfrac{x^2}{\sqrt[3]{(x^2+1)^2}+\sqrt[3]{x^2+1}+1}}{\sin x}$
$=\mathop{\lim}\limits_{x\to 0}\dfrac{\dfrac{2}{\sqrt{2x+1}+1}-\dfrac{x}{\sqrt[3]{(x^2+1)^2}+\sqrt[3]{x^2+1}+1}}{\dfrac{\sin x}{x}}=1$