Tính
$1. \lim_{x \rightarrow 0} \frac{e^{3x^2} \cos^5 x-1}{x^2}$
$2. \lim_{x \rightarrow 0} \frac{(\sqrt[3]{1+x^2}+2x)^{\frac{7}{5}}-(\sqrt[3]{1+x^2}-x)^{\frac{7}{5}}}{x}$
P/s: dùng đại lượng $VCB$.
1. Ta có
$$e^{3x^2}\sim 1+3x^2$$
$$\cos^5 x \sim \left ( 1-\frac{x^2}{2} \right )^5\sim 1-\frac{5x^2}{2}$$
$$\Rightarrow e^{3x^2}\cos^5x-1\sim \left ( 1+3x^2 \right )\left ( 1-\frac{5x^2}{2} \right )-1\sim\frac{x^2}{2}$$
Vậy $$\lim_{x \rightarrow 0} \frac{e^{3x^2} \cos^5 x-1}{x^2}=\frac{1}{2}$$
2. Tương tự, ta cũng có
$$\left ( \sqrt[3]{1+x^2}+2x \right )^{\frac{7}{5}}\sim \left ( 1+2x \right )^{\frac{7}{5}}\sim 1+\frac{14x}{5}$$
$$\left ( \sqrt[3]{1+x^2}-x \right )^{\frac{7}{5}}\sim \left ( 1-x \right )^{\frac{7}{5}}\sim 1-\frac{7x}{5}$$
$$\left ( \sqrt[3]{1+x^2}+2x \right )^{\frac{7}{5}}-\left ( \sqrt[3]{1+x^2}-x \right )^{\frac{7}{5}}\sim \left ( 1+\frac{14x}{5} \right )-\left ( 1-\frac{7x}{5} \right )=\frac{21x}{5}$$
Vậy $$\lim_{x\to 0}\frac{\left ( \sqrt[3]{1+x^2}+2x \right )^{\frac{7}{5}}-\left ( \sqrt[3]{1+x^2}-x \right )^{\frac{7}{5}}}{x}=\frac{21}{5}$$