# 什么时候该舍去较小的无穷大？以 $\lim_{x \rightarrow \infty}$ $\sin \pi \sqrt{4n^{2} + n}$ 为例

## 一、题目

$$\lim_{x \rightarrow \infty} \sin \pi \sqrt{4n^{2} + n} = ?$$

## 二、解析

$$\sqrt{4n^{2} + n} \sim \sqrt{4n^{2}} \sim 2n.$$

Next

$$\lim_{x \rightarrow \infty} \sin \pi \sqrt{4n^{2} + n} =$$

$$\lim_{x \rightarrow \infty} \sin 2 n \pi = 0$$

$$\lim_{x \rightarrow \infty} \sin 2 n \pi \Rightarrow 极限不存在$$

Next

$$\lim_{x \rightarrow \infty} \sin \pi \sqrt{4n^{2} + n} =$$

$$\lim_{x \rightarrow \infty} \sin \Big[ 2 n \pi + ( \pi \sqrt{4n^{2} + n} – 2n \pi) \Big] =$$

$$\lim_{x \rightarrow \infty} \sin \Big[ 2 n \pi + (\sqrt{4n^{2} + n} – 2n) \pi \Big] =$$

$$\lim_{x \rightarrow \infty} \sin (\sqrt{4n^{2} + n} – 2n) \pi =$$

$$\lim_{x \rightarrow \infty} \sin \Big[ \frac{(\sqrt{4n^{2} + n} – 2n)(\sqrt{4n^{2} + n} + 2n)}{\sqrt{4n^{2} + n} + 2n} \Big] \pi =$$

$$\lim_{x \rightarrow \infty} \sin \Big( \frac{2n + n – 2n}{\sqrt{4n^{2} + n} + 2n} \Big) \pi =$$

$$\lim_{x \rightarrow \infty} \sin \Big( \frac{n}{\sqrt{4n^{2} + n} + 2n} \Big) \pi =$$

Next

$$\lim_{x \rightarrow \infty} \sin \Big( \frac{n}{\sqrt{4n^{2}} + 2n} \Big) \pi =$$

$$\lim_{x \rightarrow \infty} \sin \Big( \frac{n}{4n} \Big) \pi =$$

$$\lim_{x \rightarrow \infty} \sin \frac{\pi}{4} = \frac{\sqrt 2}{2}.$$