Uniform convergence of sequence on interval $[-b,0]$ I am aiming to prove that there exists a continous function $\exp:\mathbb{R}\rightarrow\mathbb{R}$  using the sequence $$e_n^x=\sum_{k=0}^{n}\frac{x^k}{k!}$$
So far: defining the sequence $e_n^x=\sum_{k=0}^{n}\frac{x^k}{k!}$ given that this is a polynomial, I have shown that this is cauchy and therefore uniformly convergent on the reals, and taking $b\in\mathbb{Q}$ the sequence is also cauchy and thus uniformly convergent. Then $e_n^x$ is uniformly convergent for $x\in[0,b]$, I then showed that the limit function is continuous, then it must be that $\exp\colon[0,b] \rightarrow\mathbb{R}$. Now assuming that $e_n^x$ is uniformly convergent in $[0,x]$.
Now I am aiming to show that $\{e_n^{-x}\}$ is uniformly convergent on $[-b,0]$, here I am aiming to use the triangle inequality somehow so that if $exp$ is converging in $[0,b]$ then $\exp$ is converging on the union set of $\mathbb{Q}$ and $[0,b]$. After doing this it has to be that if $\exp:[0,b]\rightarrow\mathbb{R}$ then $\exp:[0,\infty]\rightarrow\mathbb{R}$. From here I can take a real number $r\in\mathbb{R}$ so that for every element $q\in\mathbb{Q}$ where $r<q$ then $\exp:[0,q]\rightarrow\mathbb{R}$ and we have that exp is continous at $r$ and this must hold for all reals.So then it must be that there exists $\exp:\mathbb{R}\rightarrow\mathbb{R}$.
I would like to know if me general idea is correct? Further I am having some trouble with showing that $\{e_n^{-x}\}$ is uniformly convergent on $[-b,0]$ how would we go about showing this?
 A: Showing that the sequence of functions $e_n(x) := \sum_{k = 0}^n \frac{x^k}{k!}$ is uniformly convergent on $[- b, 0]$ requires just as much work as showing that it is uniformly convergent on $[0, b]$.
In fact, I would say that the thing to aim for is showing that $e_n(x)$ is uniformly convergent on $[-b, b]$, for any $b \in [0, \infty)$. This suffices for proving that $e^x$ is continuous everywhere.
To get there, we can appeal to this standard result:

Theorem: If the sequence $\lim_{k \to \infty} \left|\frac{a_k }{ a_{k+1}} \right|$ converges, then $r := \lim_{k \to \infty} \left|\frac{a_k }{ a_{k+1}} \right|$ is the radius of convergence for the power series $\sum_{k = 0}^\infty a_k x^k$. (This is true even if $r = \infty$.)
Furthermore, if $0 \leq b < r$, then the sequence of partial sums $x \mapsto \sum_{k = 0}^n a_k x^k$ is uniformly convergent on $[-b, b]$.

In our case, $a_k = 1 / k!$ and $\lim_{k \to \infty} \left|\frac{a_k }{ a_{k+1}} \right| = \lim_{k \to \infty} | k + 1 | = \infty$, so the radius of convergence is $r = \infty$. Hence for any $b \in [0, \infty)$, $e_n(x)$ is uniformly convergent on $[-b, b]$.
A: For $x \in [-b,b]$ and $m \ge n$ we have
$$\left|e_m(x) - e_n(x)\right| =\left|\sum_{k=n+1}^m \frac{x^k}{k!}\right| \le \sum_{k=n+1}^m \frac{|x|^k}{k!} \le \sum_{k=n+1}^\infty \frac{b^k}{k!} \xrightarrow{m,n \to \infty} 0$$
uniformly in $x$ so the sequence $(e_n)_n$ is uniformly Cauchy on $[-b,b]$ and hence converges uniformly to a continuous function $\exp : [b,b] \to \Bbb{R}$.
You only need to know that the series $\sum_{k=1}^\infty \frac{b^k}{k!}$ converges, which you can prove using the ratio test:
$$\frac{\frac{b^{k+1}}{(k+1)!}}{\frac{b^k}{k!}} = \frac{b}{k+1} \xrightarrow{k\to\infty} 0.$$
