# 一点处连续与存在的区别：连续性要考虑“邻居”，存在性只需要考虑“自己”

## 二、解析

### 存在性的判断

$$f(x) = \ln \left(1+x^{\frac{2}{3}}\right) – x^{\frac{2}{3}}$$

\begin{aligned} \textcolor{springgreen}{f(0)} \\ & = \ln (1 + 0) – 0 \\ & = \textcolor{springgreen}{0} \end{aligned}

\begin{aligned} \textcolor{springgreen}{f^{\prime}(x)} \\ & = \left[ \ln \left(1+x^{\frac{2}{3}}\right) – x^{\frac{2}{3}} \right]^{\prime} \\ \\ & = \frac{1}{1+x^{\frac{2}{3}}} \cdot \frac{2}{3 \sqrt[3]{x}} – \frac{2}{3 \sqrt[3]{x}} \\ \\ & = \textcolor{springgreen}{\frac{2}{3 \sqrt[3]{x}} \left( \frac{1}{1+x^{\frac{2}{3}}} – 1 \right)} \end{aligned}

\begin{aligned} \textcolor{springgreen}{f^{\prime}(0)} \\ \\ & = \lim \limits_{x \rightarrow 0} \frac{f(x)-f(0)}{x – 0} \\ \\ & = \lim \limits_{x \rightarrow 0} \frac{\ln \left(1+x^{\frac{2}{3}}\right)-x^{\frac{2}{3}}}{x} \\ \\ & \underrightarrow{\text{洛必达运算 \ }} \lim \limits_{x \rightarrow 0} \frac{2}{3 \sqrt[3]{x}}\left(\frac{1}{1+x^{\frac{2}{3}}}-1\right) \\ \\ & = \lim \limits_{x \rightarrow 0} \frac{2}{3 x^{\frac{1}{3}}} \cdot \frac{\textcolor{red}{-x^{\frac{2}{3}}}}{1+x^{\frac{2}{3}}} \\ \\ & = \textcolor{red}{\frac{-2}{3}} \lim \limits_{x \rightarrow 0} \frac{x^{\frac{1}{3}}}{1+x^{\frac{2}{3}}} \\ \\ & = \textcolor{red}{\frac{-2}{3}} \cdot \frac{0}{1 + 0} \\ \\ & = \textcolor{springgreen}{0} \end{aligned}

\begin{aligned} f^{\prime \prime}(0) \\ \\ & = \lim \limits_{x \rightarrow 0} \frac{f^{\prime}(x)-f^{\prime}(0)}{x – 0} \\ \\ & = \lim \limits_{x \rightarrow 0} \frac{\frac{2}{3 \sqrt[3]{x}} \left( \frac{1}{1+x^{\frac{2}{3}}} – 1 \right)}{x} \\ \\ & = \lim \limits_{x \rightarrow 0} \frac{\frac{-2}{3} x^{\frac{1}{3}} \cdot \frac{1}{1+x^{\frac{2}{3}}}}{x} \\ \\ & = \frac{-2}{3} \lim \limits_{x \rightarrow 0} \frac{x^{\frac{1}{3}}}{1+x^{\frac{2}{3}}} \cdot \frac{1}{x} \\ \\ & = \frac{-2}{3} \lim \limits_{x \rightarrow 0} \frac{x^{\frac{1}{3}} \cdot x^{\frac{2}{3}}}{x^{\frac{2}{3}}\left( 1+x^{\frac{2}{3}} \right)} \cdot \frac{1}{x} \\ \\ & = \frac{-2}{3} \lim \limits_{x \rightarrow 0} \frac{1}{x^{\frac{2}{3}}\left(1+x^{\frac{2}{3}}\right)} \\ \\ & = \frac{-2}{3} \cdot \frac{1}{0} \\ \\ & = \infty \end{aligned}

### 连续性的判断

\begin{aligned} \lim \limits_{x \rightarrow 0} f^{\prime}(x) \\ \\ & = \lim \limits_{x \rightarrow 0} \frac{2}{3 \sqrt[3]{x}}\left(\frac{1}{1+x^{\frac{2}{3}}}-1\right) \\ \\ & = \frac{-2}{3} \lim \limits_{x \rightarrow 0} \frac{x^{\frac{1}{3}}}{1+x^{\frac{2}{3}}} \\ \\ & = \frac{-2}{3} \cdot \frac{0}{1+0} \\ \\ & = 0 \\ \\ & = f^{\prime}(0) \end{aligned}