Use Taylor series to compute $\lim\limits_{x \to 0} \left( \frac{1}{\sin(x)}- \frac{1}{x} \right) $

Question

Use Taylor series to solve $$\lim\limits_{x \to 0} \left( \frac{1}{\sin(x)}- \frac{1}{x} \right) $$

This equals $$\lim\limits_{x \to 0} \left( \sum_{n=0}^{\infty}\frac{(2n+1)!}{x^{2n+1}(-1)^{n+1}} - \sum_{n=0}^{\infty}\frac{1}{(-x+1)^n} \right)$$

I am unsure how to proceed. I tried writing out the first few terms, but nothing seemed to cancel.

Answer 1

I'm not quite sure how you got your formula. Using just the first $2$ terms of the Taylor series for $\sin(x)$ (you actually only need just $1$ term, as shown in some of the other answers, but I thought using $2$ may help to better see what is going on), such as given in Trigonometric functions, and as suggested in copper.hat's question comment, you have

$$\begin{equation}\begin{aligned} \lim_{x\to 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right) & = \lim_{x\to 0}\left(\frac{x - \sin(x)}{x\sin(x)}\right) \\ & = \lim_{x\to 0}\left(\frac{x - \left(x - \frac{x^3}{6} + O(x^5)\right)}{x\left(x - \frac{x^3}{6} + O(x^5)\right)}\right) \\ & = \lim_{x\to 0}\left(\frac{\frac{x^3}{6} - O(x^5)}{x^2 - \frac{x^4}{6} + O(x^6)}\right) \\ & = \lim_{x\to 0}\left(\frac{\frac{x}{6} - O(x^3)}{1 - \frac{x^2}{6} + O(x^4)}\right) \\ & = \frac{0}{1} \\ & = 0 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Answer 2

I strongly believe that you don't have to use the WHOLE Taylor Series, but just the firs terms.

You know that

$$\sin(x) \approx x + O(x^3)$$ as $x\to 0$ hence just substitute in the $\sin(x)$ term to get:

$$\lim_{x\to 0} \left(\frac{1}{x} - \frac{1}{x}\right) = 0$$

Answer 3

Let $ x $ be a real from $ \mathbb{R}^{*} \cdot $

Since $ \sin{x}=\sum\limits_{n=0}^{+\infty}{\left(-1\right)^{n}\frac{x^{2n+1}}{\left(2n+1\right)!}} $, we get that $ \frac{x-\sin{x}}{x^{3}}=\sum\limits_{n=0}^{+\infty}{\left(-1\right)^{n}\frac{x^{2n}}{\left(2n+3\right)!}}=\frac{1}{6}+x^{2}\sum\limits_{n=1}^{+\infty}{\left(-1\right)^{n}\frac{x^{2n-2}}{\left(2n+3\right)!}}\underset{x\to 0}{\longrightarrow}\frac{1}{6} \cdot $

Which means \begin{aligned} \lim_{x\to 0}{\left(\frac{1}{\sin{x}}-\frac{1}{x}\right)}&=\lim_{x\to 0}{x\left(\frac{x-\sin{x}}{x^{3}}\right)\left(\frac{x}{\sin{x}}\right)}\\ &=0\times\frac{1}{6}\times 1\\ \lim_{x\to 0}{\left(\frac{1}{\sin{x}}-\frac{1}{x}\right)}&=0 \end{aligned}

Answer 4

For $x\to 0$, we have $$\begin{align} \frac{1}{\sin x}=\frac{1}{x+O(x^3)}=\frac 1x\cdot\frac{1}{1+O(x^2)}=\frac 1x\left(1+O(x^2)\right)=\frac 1x+O(x) \end{align}$$ Thus, $$\frac{1}{\sin x}-\frac 1x=O(x)\;\;(x\to 0) $$ and the desired limit is $0$.