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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?

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    $\begingroup$ Once you prove that $e^{a+b}=e^{a}e^{b}$, the proof is very simple. $\endgroup$ – egreg Jun 15 '16 at 15:33
  • $\begingroup$ @egreg I'd proof it like that: e^a * e^b = exp(aln e)*exp(bln e) = exp((a+b)*ln e) = e^(a+b). But how could I use this proof for the proof I'm looking for? ^^ $\endgroup$ – tenepolis Jun 15 '16 at 15:48
  • $\begingroup$ What you have done works for showing continuity of $\exp(x)$ at $x = 0$ and your approach has no problem. And then you can follow as mentioned in egreg's answer. $\endgroup$ – Paramanand Singh Jun 17 '16 at 15:26
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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 $0$.

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  • $\begingroup$ Ah alright now I also understand why you referred to the proof of $e^{a+b} = e^{a} e^{b}$. But what does $h$ mean here? I don't know what to do with the result. $\endgroup$ – tenepolis Jun 17 '16 at 15:52
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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


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 $x$.

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  • $\begingroup$ Interesting, but the exponential is defined differently for the OP. $\endgroup$ – egreg Jun 15 '16 at 22:16
  • $\begingroup$ @egreg Well, that is the reason for my writing the introduction in which I stated "I thought it might be instructive to show that $e^x e^y=e^{x+y}$ directly from the limit definition of the exponential function ..." And this question ties to one asked earlier by the OP in which the OP mentions use of the limit definition. -Mark $\endgroup$ – Mark Viola Jun 15 '16 at 22:19
  • $\begingroup$ Why was this down voted?? $\endgroup$ – Mark Viola Jun 16 '16 at 3:24
  • $\begingroup$ I didn't downvote; however, this is in no way an answer to the question. The OP's definition of the exponential is different from the one you're using and it's by no means obvious the two ways are equivalent. $\endgroup$ – egreg Jun 16 '16 at 6:41
  • $\begingroup$ @tenepolis Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Jun 17 '16 at 14:43

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