一、题目![题目 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/f68a9e590526998388b0f9b71bd5d3f73dda4ed9764819fe8f36488fa537e9b9499f465fd201d7c117b8901c3ad071915a34a688058a739ebc39835753a8d7cc.svg)
$$
I = \lim _{ x \rightarrow \infty } \left( 2 x ^ { 3 } – x + 1 \right) = ?
$$
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二、解析 ![解析 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/6fff698aa5c66c6c7a143e3d2a00fa8ee7eab76be5360d89eb43a03143848e8cd60377c76bf830c93ec6603be5af661d9c52238834792ea548bf14de10b05ad9.svg)
首先:
$$
\begin{aligned}
& \lim _{ x \rightarrow \infty } \frac { 1 } { 2 x ^ { 3 } – x + 1 } \\ \\
= & \lim _{ x \rightarrow \infty } \frac { \frac { 1 } { x ^ { 3 } } } { 2 – \frac { 1 } { x ^ { 2 } } + \frac { 1 } { x ^ { 3 } } } \\ \\
= & \frac { \lim _{ x \rightarrow \infty } \frac { 1 } { x ^ { 3 } } } { \lim _{ x \rightarrow \infty } \left( 2 – \frac { 1 } { x ^ { 2 } } + \frac { 1 } { x ^ { 3 } } \right) } \\ \\
= & \frac{0}{2} \\ \\
= & 0
\end{aligned}
$$
于是:
$$
I = \lim_{ x \rightarrow \infty } \left( 2 x ^ { 3 } – x + 1 \right) = \infty
$$