Limit Evaluation using Taylor Polynomials.

Question

Consider the following limit problem:$$\lim \limits _{x\to 0}\frac{e^x-1}{x}.$$It's known that$$e^x=\lim \limits _{n\to \infty}\sum \limits _{r=0}^n\frac{x^r}{r!}\qquad \forall x\in N_0,$$where $N_0$ is a deleted neighbourhood of $0$.

So the original limit gets converted into the following:$$\lim \limits _{x\to 0}\frac{\lim \limits _{n\to \infty}\sum \limits _{r=0}^n\frac{x^r}{r!}-1}{x},$$which simplifies to:\begin{align*}\lim \limits _{x\to 0}\frac{\lim \limits _{n\to \infty}\left (1+\sum \limits _{r=1}^n\frac{x^r}{r!}\right )-1}{x} & =\lim \limits _{x\to 0}\frac{\lim \limits _{n\to \infty}\sum \limits _{r=1}^n\frac{x^r}{r!}}{x} \\ & =\lim \limits _{x\to 0}\left (\lim \limits _{n\to \infty}1+\sum \limits _{r=2}^n\frac{x^{r-1}}{r!}\right ) \\ & =1. \end{align*}My only problem with the above method is how can we say that the last limit will for sure be $1$? Precisely, how are we so sure that the last summation($\displaystyle \sum \limits _{r=2}^n\frac{x^{r-1}}{r!}$) will be $0$ when both limits (of $n$ and of $x$) are applied to it?

I'm still in high school so I don't know how to evaluate limits using epsilon-delta definition. Is there any other way(other than using epsilon-delta definition) to convince myself that the last summation indeed is $0$ in the limit?

PS: I know other more efficient ways to approach the above limit problem, however I have a similar type of problem in every limit I evaluate using taylor expansions. So I'm using this limit as a placeholder for my general problem.

Answer 1

In high school? Don't worry about it. Otherwise, there are two fairly straightforward approaches.

Observation number 1: The uniform limit of continuous functions is continuous, and these types of analytic power series converge uniformly (in their radius of convergence). With this observation: $$\sum_{r=2}^\infty\frac{x^{r-1}}{r!}$$Is a continuous function. The limit as $x\to0$ is its evaluation at $0$, so just $0$ as you expect.

Approach 2: We can attempt to explicitly bound. When $|x|\lt1$, $|x|^{r-1}\le|x|$ for all $r\ge2$.

$$\left|\sum_{r=2}^\infty\frac{x^{r-1}}{r!}\right|\le\sum_{r=2}^\infty\frac{|x|^{r-1}}{r!}\le|x|\cdot\sum_{r=2}^\infty\frac{1}{r!}=|x|\cdot(e-2)$$

Letting $|x|\to0$ gives a clear limit of zero, using this bound.