Limit of a rational expression involving sinus

Question

3.72K 浏览2021年3月23日mathematics

0

admin 166 2021年3月23日 0条评论

The task is to find
$L=\lim _{n \rightarrow \infty} n^{2}-\frac{n}{\sin \left(\frac{1}{n}\right)}$
For sufficiently large $n$ we have $\sin \left(\frac{1}{n}\right) \approx \frac{1}{n}$ so that
$L \approx n^{2}-\frac{n}{\frac{1}{n}}=n^{2}-n^{2}=0$
so the limit $L$ is zero. Am I correct?

admin 更改状态以发布 2021年3月23日

1 答案

admin · Answer 1 · 2021-03-23T14:24:30+00:00

$\sin \left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{6 n^{3}}+\frac{1}{120 n^{5}}+O\left(\frac{1}{n^{7}}\right)$
Now long division
$\begin{array}{l} \frac{1}{\sin \left(\frac{1}{n}\right)}=n+\frac{1}{6 n}+\frac{7}{360 n^{3}}+O\left(\frac{1}{n^{5}}\right) \\ n^{2}-\frac{n}{\sin \left(\frac{1}{n}\right)}=-\frac{1}{6}-\frac{7}{360 n^{2}}+O\left(\frac{1}{n^{4}}\right) \end{array}$
which shows the limit and how it is approached.

Limit of a rational expression involving sinus

Published by admin on 2021年3月23日2021年3月23日

1 答案