[例1] (2004) $\lim _{x \rightarrow 0}\left(\frac{1}{\sin ^{2} x}-\frac{\cos ^{2} x}{x^{2}}\right)=\_\_. $

$\lim _{x \rightarrow \infty} x^{2}\left(2-x \sin \frac{1}{x}-\cos \frac{1}{x}\r

$\lim _{x \rightarrow 0}(1-\sin 2x)^{\frac{1}{x}}=\_\_\_\_\_\_.$ $\lim _{x \rig

$\lim _{x \rightarrow 0^{-}} e^{\frac{1}{x}}=\_\_\_\_\_\_\_\_.$ $\lim _{x \righ

$\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}

求极限 $\lim_{x \to 0} \frac{e^x \sin x - x(x+1)}{\sin^3 x}$. 求极限 $\lim_{x \to 0}

求极限 $\lim _{x \rightarrow +\infty}\left(x+e^{x}\right)^{\frac{1}{x}}$. 求极限 $\li

求极限 $\lim_{x \to 0} \frac{\int_{0}^{x} \sin t^{2} dt}{x - \arctan x}$. 求极限 $\li

$\lim_{{x \to \infty}} (\sqrt[3]{x^3 + x^2} - xe^{\frac{1}{x}}) = \_\_\_\_\_\_.$

已知当 $x \to 0$ 时, $x^2 \ln \left(1 + x^2\right)$ 是 $\sin^n x$ 的高阶无穷小, 而 $\sin^n x

求极限$\lim _{n \infty}\left(\frac{1}{n^{2}+n+1}+\frac{2}{n^{2}+n+2}+\cdots+\frac{n