Proofing that the exponential function is continuous in every $x_{0}$

Question

Given:

$$\exp: \mathbb{R} \ni x \mapsto \sum_{k=0}^{\infty } \frac{1}{k!} x^{k} \in \mathbb{R}$$

also $e = \exp(1)$. For all $x \in \mathbb{R}$ with $\left | x \right | \leq 1$: $$\left | \exp(x) - 1 \right | \leq \left | x \right | \cdot (e-1)$$

and $\exp(0) = 1$

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In order to proof that the exponential function is continuous for every $x_{0}$, it needs to be shown that it's continuous at all. This was shown here (it's continuous at $x_{0}$ = 0): Proving that the exponential function is continuous

But I prefer this proof:

$$ \exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!} $$ apply some little changes $$ \exp(x) = 1 + x \sum_{n=1}^\infty \frac{x^{n-1}}{n!} $$ whence for $x\to 0$ $$ |\exp(x) - 1| \le |x| \sum_{n=1}^\infty{|x|^{n-1}} \le |x| \frac{1}{1-|x|} \to 0 $$

I would say in order to show that the exponential function is continuous for all $x_{0} = 0$, I just need to show it is continuous at $x_{0}$ = 0 (done) and then I can just conclude it is continuous everywhere, so at $x=x_{0}$? Not sure about this, is it really possible?

Answer 1

The proof that $$ \exp(a+b)=\exp(a)\exp(b) $$ is a simple application of absolute convergence and binomial theorem. See Prove $e^{x+y}=e^{x}e^{y}$ by using Exponential Series

Once you have this knowledge, you can observe that $$ |\exp(x+h)-\exp(x)|=|\exp(x)|\,|\exp(h)-1| $$ and use the already proved continuity at $0$.

Answer 2

I thought it might be instructive to show that $e^xe^y=e^{x+y}$ directly from the limit definition of the exponential function given by

$$\bbox[5px,border:2px solid #C0A000]{e^x=\lim_{n\to \infty}\left(1+\frac{x}{n}\right)^n} \tag 1$$

To that end, we now proceed

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First, we see that

$$\begin{align} \left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n&=\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^n\\\\ &=\left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n \tag 2 \end{align}$$

If $xy>0$, then for all $n>N$ we have from $(2)$

$$\left(1+\frac{x+y}{n}\right)^n\le \left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n \le \left(1+\frac{x+y+\frac{xy}{N}}{n}\right)^n \tag 3$$

Taking the limit as $n\to \infty$ of $(3)$ reveals

$$e^{x+y}\le e^{x}e^{y} \le e^{x+y+xy/N} \tag 4$$

Using the left-hand side inequality in $(4)$ and applying it to the right-hand side shows that for $xy>0$, $e^{x+y+xy/N}\le e^{x+y}e^{xy/N}$ and therefore

$$e^{x+y}\le e^{x}e^{y} \le e^{x+y} e^{xy/N} \tag 5$$

Letting $N\to \infty$ yields

$$e^{x+y}\le e^xe^y\le e^{x+y}$$

where we used the inequalities for the exponential $1+x\le e^x\le \frac{1}{1-x}$ for $x<1$ that I established in THIS ANSWER using only $(1)$ and Bernoulli's Inequality.

We have now established the equality $e^xe^y=e^{x+y}$ for $xy>0$. For $xy<0$, we simply reverse the inequalities in $(3)$ and proceed analogously.

Finally, refer to THIS ANSWER in which the result herein was used to prove the continuity of $e^x$ for all $x$.