# 集火攻击：多种方法解一道题

## 一、题目

$$\lim \limits_{x \rightarrow 0} \frac{\left(3+\sin x^{2}\right)^{x}-3^{\sin x}}{x^{3}} = ?$$

## 二、解析

### 方法 1

$$\lim \limits_{x \rightarrow 0} \frac{\left(3+\sin x^{2}\right)^{x}-3^{\sin x}}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{\left(3+\sin x^{2}\right)^{x}-3^{x}+3^{x}-3^{\sin x}}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{\left(3+\sin x^{2}\right)^{x}-3^{x}}{x^{3}}+\lim \limits_{x \rightarrow 0} \frac{3^{x}-3^{\sin x}}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{\left[3\left(1+\frac{\sin x^{2}}{3}\right)\right]^{x}-3^{x}}{x^{3}}+\lim \limits_{x \rightarrow 0} \frac{\left(3^{\xi}\right)^{\prime}(x-\sin x)}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{3^{x}\left(1+\frac{\sin x^{2}}{3}\right)^{x}-3^{x}}{x^{3}}+\lim \limits_{x \rightarrow 0} \frac{3^{\xi} \ln 3(x-\sin x)}{x^{3}}$$

$3^{x}$ $=$ $3^{\xi}$ $=$ $1$, 可以直接写成常数或舍去：

$$\lim \limits_{x \rightarrow 0} \frac{\left(1+\frac{\sin x^{2}}{3}\right)^{x}-1}{x^{3}}+\lim \limits_{x \rightarrow 0} \frac{\ln 3(x-\sin x)}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{\frac{x \sin x^{2}}{3}}{x^{3}}+\lim \limits_{x \rightarrow 0} \frac{\ln 3 \cdot \frac{1}{6} x^{3}}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{\frac{x^{3}}{3}}{x^{3}}+\frac{1}{6} \ln 3=\frac{1}{3}+\frac{1}{6} \ln 3.$$

### 方法 2

$$\lim \limits_{x \rightarrow 0} \frac{\left(3+\sin x^{2}\right)^{x}-3^{\sin x}}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{e^{x \ln \left(3+\sin x^{2}\right)}-e^{\sin x \ln 3}}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{\left[e^{x \ln \left(3+\sin x^{2}\right)}-1\right]-\left[e^{\sin x \ln ^{3}}-1\right]}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{x \ln \left(3+\sin x^{2}\right)-\sin x \ln 3}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{\ln \left(3+\sin x^{2}\right)+x \frac{2 x \cos x^{2}}{3+\sin x^{2}}-\cos x \ln 3}{3 x^{2}}=$$

$$\lim \limits_{x \rightarrow 0}\left[\frac{\ln \left(3+\sin x^{2}\right)}{3 x^{2}}+\frac{2 x^{2} \cos x^{2}}{3 x^{2}\left(3+\sin x^{2}\right)}-\frac{\cos x \ln 3}{3 x^{2}}\right]=$$

$$\lim \limits_{x \rightarrow 0}\left[\frac{\frac{2 x \cos x^{2}}{3+\sin x^{2}}}{6 x}+\frac{2}{3\left(3+\sin x^{2}\right)}-\frac{-\sin x \ln 3}{6 x}\right]=$$

$$\lim \limits_{x \rightarrow 0}\left[\frac{2}{18}+\frac{2}{9}+\frac{1}{6} \ln 3\right]=\frac{1}{3}+\frac{1}{6} \ln 3.$$

### 方法 3

$$\lim \limits_{x \rightarrow 0} [e^{x \ln \left(3+\sin x^{2}\right)}-e^{\sin x \ln 3} ]=$$

$$\lim \limits_{x \rightarrow 0} e^{\xi}\left[x \ln \left(3+\sin x^{2}\right)-\sin x \ln 3\right] \Rightarrow$$

$$\lim \limits_{x \rightarrow 0} [x \ln \left(3+\sin x^{2}\right)-\sin x \ln 3].$$

### 方法 4

$$\lim \limits_{x \rightarrow 0} [e^{x \ln \left(3+\sin x^{2}\right)}-e^{\sin x \ln 3} ]=$$

$$\lim \limits_{x \rightarrow 0} e^{\sin x \ln 3}\left[e^{x \ln \left(3+\sin x^{2}\right)-\sin x \ln 3}-1\right]=$$

$$\lim \limits_{x \rightarrow 0} e^{\sin x \ln 3}\left[x \ln \left(3+\sin x^{2}\right)-\sin x \ln 3\right]=$$

$\lim \limits_{x \rightarrow 0} e^{\sin x \ln 3}$ $=$ $1$ $\Rightarrow$

$$\lim \limits_{x \rightarrow 0} x \ln \left(3+\sin x^{2}-\sin x \ln 3\right).$$

### 方法 5

$$\lim \limits_{x \rightarrow 0} \frac{x \ln \left(3+\sin x^{2}\right)-\sin x \ln 3}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} (x-\sin x \sim \frac{1}{6} x^{3} ) \Rightarrow \lim \limits_{x \rightarrow 0} (\sin x \sim x-\frac{1}{6} x^{3} ) \Rightarrow$$

$$\lim \limits_{x \rightarrow 0} \frac{x \ln \left(3+\sin x^{2}\right)-\left(x-\frac{1}{6} x^{3}\right) \ln 3}{x^{3}} \Rightarrow$$

$$\ln \left(3+\sin x^{2}\right)=\ln \left[3\left(1+\frac{\sin x^{2}}{3}\right)\right]=$$

$$\ln 3+\ln \left(1+\frac{\sin x^{2}}{3}\right).$$

$$\lim \limits_{x \rightarrow 0} \frac{x \ln 3+x \ln \left(1+\frac{\sin x^{2}}{3}\right)-x \ln 3+\frac{x^{3}}{6} \ln 3}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{x \ln \left(1+\frac{\sin x^{2}}{3}\right)}{x^{3}}+\lim \limits_{x \rightarrow 0} \frac{\frac{x^{3}}{6} \ln 3}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{\frac{x \sin x^{2}}{3}}{x^{3}}+\lim \limits_{x \rightarrow 0} \frac{x^{3}}{6 x^{3}} \ln 3=$$

$$\frac{1}{3}+\frac{1}{6} \ln 3.$$

### 方法 6

$$\lim \limits_{x \rightarrow 0} \frac{x \ln \left(3+\sin x^{2}\right)-\sin x \ln 3}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{x\left[\ln 3\left(1+\frac{\sin x^{2}}{3}\right)\right]-\sin x \ln 3}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{x \ln 3+x \ln \left(1+\frac{\sin x^{2}}{3}\right)-\sin x \ln 3}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{x \ln \left(1+\frac{\sin x^{2}}{3}\right)+(x \ln 3-\sin x \ln 3)}{x^{3}}=$$

$$\lim \limits_{x \rightarrow 0} \frac{x \ln \left(1+\frac{\sin x^{2}}{3}\right)}{x^{3}}+\lim \limits_{x \rightarrow 0} \ln 3 \cdot \frac{x-\sin x}{x^{3}} =$$

$$\lim \limits_{x \rightarrow 0} \frac{x \cdot \frac{x^{2}}{3}}{x^{3}}+\lim \limits_{x \rightarrow 0} \ln 3 \cdot \frac{\frac{1}{6} x^{3}}{x^{3}}=$$

$$\frac{1}{3}+\frac{1}{6} \ln 3.$$