一、题目
$$
\lim_{x \rightarrow \infty} \Big[ \frac{x^{2}}{(x – a) (x + b)} \Big]^{x} = ?
$$
难度评级：
二、解析 
$$
\lim_{x \rightarrow \infty} \Big[ \frac{x^{2}}{(x – a) (x + b)} \Big]^{x} =
$$
$$
\lim_{x \rightarrow \infty} \Big( \frac{x}{x – a} \cdot \frac{x}{x + b} \Big)^{x} =
$$
$$
\lim_{x \rightarrow \infty} \Big( \frac{x}{x – a}\Big)^{x} \cdot \Big(\frac{x}{x + b} \Big)^{x} =
$$
Next 
$$
\lim_{x \rightarrow \infty} \Big[ 1 + \big(\frac{x}{x – a} – 1 \big) \Big]^{x} \cdot \Big[ 1 + \big( \frac{x}{x + b} – 1 \big) \Big]^{x} =
$$
$$
\lim_{x \rightarrow \infty} \Big( 1 + \frac{a}{x – a} \Big)^{x} \cdot \Big( 1 + \frac{-b}{x + b} \Big)^{x} =
$$
$$
\lim_{x \rightarrow \infty} \Big( 1 + \frac{a}{x – a} \Big)^{\frac{x-a}{a} \cdot \frac{ax}{x-a}} \cdot \Big( 1 + \frac{-b}{x + b} \Big)^{\frac{x+b}{-b} \cdot \frac{-bx}{x+b}} =
$$
$$
\lim_{x \rightarrow \infty} e^{\frac{ax}{x-a}} \cdot e^{\frac{-bx}{x+b}} =
$$
$$
\lim_{x \rightarrow \infty} e^{\frac{ax}{x-a} + \frac{-bx}{x+b}} =
$$
$$
\lim_{x \rightarrow \infty} e^{\frac{ax}{x} + \frac{-bx}{x}} =
$$
$$
\lim_{x \rightarrow \infty} e^{a-b}.
$$
Next
方法二
$$
\lim_{x \rightarrow \infty} \Big[ \frac{x^{2}}{(x – a) (x + b)} \Big]^{x} =
$$
$$
\lim_{x \rightarrow \infty} \Big( \frac{x}{x – a} \cdot \frac{x}{x + b} \Big)^{x} =
$$
$$
\lim_{x \rightarrow \infty} \Big( \frac{x}{x – a}\Big)^{x} \cdot \Big(\frac{x}{x + b} \Big)^{x} =
$$
Next
$$
\lim_{x \rightarrow \infty} \Big( \frac{x – a}{x}\Big)^{-x} \cdot \Big(\frac{x + b}{x} \Big)^{-x} =
$$
$$
\lim_{x \rightarrow \infty} \Big( 1 – \frac{a}{x}\Big)^{-x} \cdot \Big(1 + \frac{b}{x} \Big)^{-x} =
$$
$$
\lim_{x \rightarrow \infty} \Big( 1 – \frac{a}{x}\Big)^{\frac{x}{-a} \cdot \frac{ax}{x}} \cdot \Big(1 + \frac{b}{x} \Big)^{\frac{x}{b} \cdot \frac{-bx}{x}} =
$$
$$
\lim_{x \rightarrow \infty} e^{\frac{ax}{x}} \cdot e^{\frac{-bx}{x}} =
$$
$$
\lim_{x \rightarrow \infty} e^{a} \cdot e^{-b} =
$$
$$
\lim_{x \rightarrow \infty} e^{a – b}.
$$