mathsbg

Tìm giới hạn của
$\lim_{x\rightarrow 0}\frac{\sqrt[3]{3x^2-1}+\sqrt{2x^2+1}}{1-\cos2x}$.

$\lim_{x\rightarrow 0}\frac{\sqrt[3]{3x^2-1}+\sqrt{2x^2+1}}{1-\cos2x}=\lim_{x\rightarrow 0}\frac{\left(\sqrt[3]{3x^2-1}+1\right)+\left(\sqrt{2x^2+1}-1\right)}{2\sin^2 x}=\lim_{x\rightarrow 0}\frac{3x^2-1+1}{2\sin^2 x\left(\sqrt[3]{(3x^2-1)^2}-\sqrt[3]{3x^2-1}+1\right)}+\lim_{x\rightarrow 0}\dfrac{2x^2+1-1}{2\sin^2 x\left(\sqrt{2x^2+1}+1\right)}=1$
Vì $\lim_{x\rightarrow 0}\dfrac{\sin x}{x}=1$.