关于幂指函数的无穷大比较的一个重要结论

一、题目

$$
I = \lim \limits_{n \rightarrow \infty} \frac{5^{n}+2^{n}}{5^{n+1}+2^{n+1}} = ?
$$

难度评级：

二、解析

$$
\begin{aligned}
I = \\ \\
& \lim \limits_{n \rightarrow \infty} \frac{5^{n}+2^{n}}{5^{n+1}+2^{n+1}} = \\ \\
& \lim \limits_{n \rightarrow \infty} \frac{5^{n}\left(1+\frac{2^{n}}{5^{n}}\right)}{5^{n+1}\left(1+\frac{2^{n+1}}{5^{n+1}}\right)} = \\ \\
& \lim \limits_{n \rightarrow \infty} \frac{5^{n}\left[1+\left(\frac{2}{5}\right)^{n}\right]}{5^{n+1}\left[1+\left(\frac{2}{5}\right)^{n+1}\right]} = \\ \\
& \lim \limits_{n \rightarrow \infty} \frac{5^{n}(1+0)}{5^{n+1}(1+0)}= \\ \\
& \lim \limits_{n \rightarrow \infty} \frac{5^{n}}{5^{n+1}} = \\ \\
& \lim \limits_{n \rightarrow \infty} \frac{5^{n}}{5^{n} \cdot 5}=\frac{1}{5}
\end{aligned}
$$

三、结论

如果 $a > b$, 则：

$$
\lim \limits_{n \rightarrow \infty} ( a^{n} + b^{n} ) = \lim \limits_{n \rightarrow \infty} a^{n} = \infty
$$

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