<s>[INST] Use Taylor series to compute $\lim\limits_{x \to 0} \left( \frac{1}{\sin(x)}- \frac{1}{x} \right) $ 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.
 [/INST] 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
$$\begin{equation}\begin{aligned}
\lim_{x\to 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right) & = \lim_{x\to 0}\left(\frac{x - \sin(x)}{x\sin(x)}\right) \\
& = \lim_{x\to 0}\left(\frac{x - \left(x - \frac{x^3}{6} + O(x^5)\right)}{x\left(x - \frac{x^3}{6} + O(x^5)\right)}\right) \\
& = \lim_{x\to 0}\left(\frac{\frac{x^3}{6} - O(x^5)}{x^2 - \frac{x^4}{6} + O(x^6)}\right) \\
& = \lim_{x\to 0}\left(\frac{\frac{x}{6} - O(x^3)}{1 - \frac{x^2}{6} + O(x^4)}\right) \\
& = \frac{0}{1} \\
& = 0
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
</s>