$\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right) \cdots\left(1+\frac{n}{n}\right)\right]^{\frac{1}{n}}=\_\_\_\_\_\_.$
$\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right) \cdots\left(1+\frac{n}{n}\right)\right]^{\frac{1}{n}}=\_\_\_\_\_\_.$
求极限$\lim _{n \infty}\left(\frac{1}{n^{2}+n+1}+\frac{2}{n^{2}+n+2}+\cdots+\frac{n
计算下列极限:(11) $\lim_{n \to \infty} \left(1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}
$\lim _{x \rightarrow \infty} x^{2}\left(2-x \sin \frac{1}{x}-\cos \frac{1}{x}\r
求极限 $\lim _{x \rightarrow +\infty}\left(x+e^{x}\right)^{\frac{1}{x}}$. 求极限 $\li
[例1] (2004) $\lim _{x \rightarrow 0}\left(\frac{1}{\sin ^{2} x}-\frac{\cos ^{2}
$\lim _{x \rightarrow 0^{-}} e^{\frac{1}{x}}=\_\_\_\_\_\_\_\_.$ $\lim _{x \righ
已知 $f(x) = \lim_{n \to \infty} \frac{\ln(e^n + x^n)}{n}$, $(x > 0)$.(1) 求 $f(x)$
设函数 $f(x) = \lim_{n \to \infty} \frac{1 + x}{1 + x^{2n}}$,讨论函数 $f(x)$ 的间断点,其结论为(
计算:(1) $87 \times \left(-\frac{5}{29} - \frac{2}{3}\right)$;(2) $(-60) \times \l
$\lim _{x \rightarrow 0}(1-\sin 2x)^{\frac{1}{x}}=\_\_\_\_\_\_.$ $\lim _{x \rig