<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] 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}
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