# 只有因“极限变量”导致的极限取值不同才叫极限不存在：因式子中其他变量取值不同导致的极限不同只能表现为“分段式极限存在”

## 一、题目

(1) $\lim \limits_{x \rightarrow \infty}\left(5 x^{5}-3 x^{3}+2\right)$.

(2) $\lim \limits_{x \rightarrow 0} \frac{1}{x^{2}} \sin \frac{1}{x}$.

(3) 数列极限 $I=\lim \limits_{n \rightarrow \infty} \sqrt[n]{\frac{a^{n}}{1+a^{n}}}$, 其中常数 $a>0$.

(4) $\lim \limits_{x \rightarrow+\infty}\left(1+\frac{1}{x}\right)^{x^{2}}$.

## 二、解析

### (1)

$$\lim \limits_{x \rightarrow \infty}\left(5 x^{5}-3 x^{3}+2\right)=$$

$$\lim \limits_{x \rightarrow \infty} x^{5}\left(5+\frac{3}{x^{2}}+\frac{2}{x^{5}}\right) =$$

$$\lim \limits_{x \rightarrow \infty} 5 x^{5}=+\infty$$

### (2)

$\sin \frac{1}{x}$ 在 $x = 0$ 附近是有界震荡无极限，这就意味着，$\sin \frac{1}{x}$ 在 $x = 0$ 的邻域内有无数多个点等于 $1$, 也有无数多个点等于 $0$.

### (3)

$$\lim \limits_{n \rightarrow \infty} \sqrt[n]{a}=\lim \limits_{n \rightarrow \infty} a^{\frac{1}{n}}=a^{0}=1$$

$$\lim \limits_{n \rightarrow \infty} q^{n}=0, \quad(|q|<1)$$

$$I=\lim \limits_{n \rightarrow \infty} \sqrt[n]{\frac{a^{n}}{1+a^{n}}}=$$

$$\lim \limits_{n \rightarrow \infty}\left(\frac{a^{n}}{1+a^{n}}\right)^{\frac{1}{n}}=$$

$$\lim \limits_{n \rightarrow \infty} a \cdot\left(\frac{1}{1+a^{n}}\right)^{\frac{1}{n}}=a \cdot 1^{0}=a$$

$$I=\lim \limits_{n \rightarrow \infty} \sqrt[n]{\frac{a^{n}}{1+a^{n}}}=\lim \limits_{n \rightarrow \infty}\left(\frac{a^{n}}{1+a^{n}}\right)^{\frac{1}{n}}=$$

$$\lim \limits_{n \rightarrow \infty}\left(\frac{1}{\frac{1}{a^{n}}+1}\right)^{\frac{1}{n}}=\lim \limits_{n \rightarrow \infty}\left(\frac{1}{1+\left(\frac{1}{a}\right)^{n}}\right)^{\frac{1}{n}}=1$$

$$\lim \limits_{n \rightarrow \infty} \sqrt[n]{\frac{a^{n}}{1+a^{n}}}=\lim \limits_{n \rightarrow \infty}\left(\frac{1}{2}\right)^{\frac{1}{n}}=1$$

$$I=\left\{\begin{array}{ll}a, & 0<a<1 \\ 1, & a \geqslant 1\end{array}\right.$$

### (4)

$$\lim \limits_{x \rightarrow+\infty}\left(1+\frac{1}{x}\right)^{x^{2}}=\lim \limits_{x \rightarrow+\infty} e^{x^{2} \ln \left(1+\frac{1}{x}\right)}=$$

$$\lim \limits_{x \rightarrow+\infty} e^{\frac{\ln \left(1+\frac{1}{x}\right)}{\frac{1}{x} \cdot \frac{1}{x}}}=\lim \limits_{x \rightarrow+\infty} e^{\frac{\frac{1}{x}}{\frac{1}{x} \cdot \frac{1}{x}}}=$$

$$\lim \limits_{x \rightarrow+\infty} e^{x}=+\infty$$

$$\lim \limits_{x \rightarrow+\infty}\left(1+\frac{1}{x}\right)^{x^{2}}=\lim \limits_{x \rightarrow+\infty}\left[\left(1+\frac{1}{x}\right)^{x}\right]^{x}=$$

$$\lim \limits_{x \rightarrow+\infty} e^{x}=+\infty$$