# 对于不是分式的式子一般不能直接“抓大头”

## 二、解析

### 奇偶性

\begin{aligned} \textcolor{springgreen}{f(-x)} \\ & = \sqrt{1-x+x^{2}} \quad – \quad \sqrt{1+x+x^{2}} \\ \\ & = – \left( \sqrt{1+x+x^{2}} \quad – \quad \sqrt{1-x+x^{2}} \right) \\ \\ & = \textcolor{springgreen}{-f(x)} \end{aligned}

### 有界性

\begin{aligned} \lim \limits_{\textcolor{orangered}{ \boldsymbol{ x \rightarrow + \infty }}} f(x) & = \lim \limits_{x \rightarrow + \infty}\left(\sqrt{1+x+x^{2}} \quad – \quad \sqrt{1-x+x^{2}}\right) \\ \\ & = \lim \limits_{x \rightarrow + \infty}\left( \frac{\sqrt{1+x+x^{2}} \quad – \quad \sqrt{1-x+x^{2}}}{1} \right) \\ \\ & = \lim \limits_{x \rightarrow + \infty} \frac{2 x}{\sqrt{1+x+x^{2}} \quad + \quad \sqrt{1-x+x^{2}}} \\ \\ & = \lim \limits_{x \rightarrow + \infty} \frac{\sqrt{x^{2}} + \sqrt{x^{2}}}{\sqrt{1+x+x^{2}} \quad + \quad \sqrt{1-x+x^{2}}} \\ \\ & = \lim \limits_{x \rightarrow + \infty} \frac{x + x}{\sqrt{1+x+x^{2}} \quad + \quad \sqrt{1-x+x^{2}}} \\ \\ & = \lim \limits_{x \rightarrow + \infty} \frac{2x}{x \sqrt{\frac{1}{x^{2}} + \frac{1}{x}+1} \quad + \quad x \sqrt{\frac{1}{x^{2}}-\frac{1}{x}+1}} \\ \\ & = \lim \limits_{x \rightarrow + \infty} \frac{2}{\sqrt{\frac{1}{x^{2}}+\frac{1}{x}+1} \quad + \quad \sqrt{\frac{1}{x^{2}}-\frac{1}{x}+1}} \\ \\ & = \textcolor{springgreen}{1} \end{aligned}

\begin{aligned} \lim \limits_{\textcolor{orangered}{ \boldsymbol{ x \rightarrow – \infty }}} f(x) & = \lim \limits_{x \rightarrow – \infty}\left(\sqrt{1+x+x^{2}}\quad – \quad \sqrt{1-x+x^{2}}\right) \\ \\ & =\lim \limits_{x \rightarrow – \infty}\left( \frac{\sqrt{1+x+x^{2}} \quad – \quad \sqrt{1-x+x^{2}}}{1} \right) \\ \\ & = \lim \limits_{x \rightarrow – \infty} \frac{ \sqrt{x^{2}} + \sqrt{x^{2}} }{\sqrt{1+x+x^{2}} \quad + \quad \sqrt{1-x+x^{2}}} \\ \\ & = \lim \limits_{x \rightarrow – \infty} \frac{ (-x) + (-x) }{\sqrt{1+x+x^{2}} \quad + \quad \sqrt{1-x+x^{2}}} \\ \\ & = \lim \limits_{x \rightarrow – \infty} \frac{ -2x }{\sqrt{1+x+x^{2}} \quad + \quad \sqrt{1-x+x^{2}}} \\ \\ & = \lim \limits_{x \rightarrow – \infty} \frac{-2x}{x \sqrt{\frac{1}{x^{2}} + \frac{1}{x}+1} \quad + \quad x \sqrt{\frac{1}{x^{2}}-\frac{1}{x}+1}} \\ \\ & = \lim \limits_{x \rightarrow – \infty} \frac{-2}{\sqrt{\frac{1}{x^{2}}+\frac{1}{x}+1} \quad + \quad \sqrt{\frac{1}{x^{2}}-\frac{1}{x}+1}} \\ \\ & = \frac{-2}{2} \\ \\ & = \textcolor{springgreen}{-1} \end{aligned}

$$\begin{cases} \lim \limits_{x \rightarrow + \infty} f(x) = 1 \\ \\ \lim \limits_{x \rightarrow – \infty} f(x) = -1 \end{cases} \textcolor{orangered}{ \quad \nRightarrow \quad } \lim \limits_{x \rightarrow \infty} f(x) = 1$$