The process of using taylor series to evaluate limits.
Matthew Martinez 
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?
$\endgroup$ 63 Answers
$\begingroup$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
$$\lim_{x\to 0}\frac{\sin x-\left(x-\frac{x^3}{3!}\right)}{x^5}=\lim_{x\to 0}\frac{\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+O(x^{11})}{x^5}=\frac1{5!}$$
$\endgroup$ 5 $\begingroup$The limit is not $\frac{1}{5!}$, in fact the limit does not exist (or, you could say the limit is $+\infty$). You need to take the limit of each individual term. Notice that the first 2 terms diverge at $x=0$ after you divide by $x^5$.
In short, you can't just blindly ignore the other terms. That only works if all the other terms go to $0$.
$\endgroup$ 1 $\begingroup$Others have pointed out that the limit does not really exist... But, addressing the general question, it is probably easier to understand (and more correct) if you use Taylor's polynomial instead of taylor's series. Imagine you want to compute a limit that actually exists, like the well known $\lim_{x \to 0}\dfrac{\sin x}{x}$, using Taylor's formula. You know that$$ \sin x = x - \frac{\cos (\xi_x)}{3!} x^3, \quad \xi_x \in (0,x) $$
and so,
$$ \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x\to 0}\dfrac{x - \dfrac{\cos (\xi_x)}{3!} x^3}{x} = 1-\frac 16 \underbrace{\lim_{x \to 0}\cos(\xi_x)x^2}_{=0} = 1 $$
$\endgroup$
