<s>[INST] The process of using taylor series to evaluate limits. For instance we want evaluate this simple limit using taylor series :
$$L=\lim_{x\to 0}\frac{\sin x}{x^5}=\lim_{x\to 0}\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+\cdots}{x^5}$$
In this case we only care about coefficient of $x^5$ to cancel out $x^5$s in  the fraction and get $L=\frac{1}{5!}$. This is a simple example of what we usually do when evaluating limit using taylor series.
My question is why we ignore other terms of the numerator? I mean:
$$L=\lim_{x\to0}\color{red}{\frac{1}{x^4}-(\frac{1}{3!}\times\frac{1}{x^2})}+\color{green}{\frac{1}{5!}}\color{red}{-(\frac{1}{7!}\times x^2)+(\frac{1}{9!}\times x^4)+\cdots}$$
I don't understand why we can ignore other terms (showed in red color) and confirm $\frac1{5!}$ as the answer. It is obvious that terms appeared on the rightside of $\color{green}{\frac1{5!}}$ are equal to zero but what about other side?
 [/INST] We can't ignore the other terms, we obtain indeed
$$\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+\cdots}{x^5}=\frac1{x^4}-\frac1{3!x^2}+\frac1{5!}+O(x^2) \to \infty$$
the limit is finite for
$$\lim_{x\to 0}\frac{\sin x-\left(x-\frac{x^3}{3!}\right)}{x^5}=\lim_{x\to 0}\frac{\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+O(x^{11})}{x^5}=\frac1{5!}$$
</s>