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I need to show that

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

is continuous at $x_0 = 0$.


It is same as $f(x)=e^{x}$, so I will show continuity for this at $x_{0}=0$.

$\lim_{x\rightarrow 0^{-}}(e^{x})= e^{0}=1$

$\lim_{x\rightarrow 0^{+}}(e^{x})= e^{0}=1$

$f(0)=e^{0}=1$

Thus the function $f(x)=e^{x}$ is continuous at $x_{0}=0$.


I'm not sure if I was allowed to convert the "thing" above to the function $f(x)=e^{x}$. Is it correct?

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    $\begingroup$ I would say it defeats the purpose of the exercise, you say "oh yes exp is actually the exponential function" and then implicitely you use the fact that the exponential is continuous (to deduce the two limits). I think if you're given the serie definition, you have to use this definition and usual theorems about convergence, continuity and/or swapping of limits/infinite summation to deduce that $\text{lim}_{x \to 0} \sum^\infty \frac{1}{k!} x_k = 1$ $\endgroup$ – yago Aug 11 '16 at 12:46
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It is relatively easy to see by using estimates like $\frac1{(k+1)!}\le\frac1{k!}$ that $$ |\exp(x)-1|\le |x|·e^{|x|} $$ or with a geometric majorant using $\frac1{(k+1)!}\le\frac1{2^k}$ $$ |\exp(x)-1|\le |x|·\frac1{1-\frac{|x|}2}. $$

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The statement

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

is often used as the definition of $\exp(x) = e^x$. In this case, the question is to prove that $\exp$, defined as above is continuous, and you can't use any facts you know about $\exp(x)$ or $e^x$, because they haven't, strictly speaking, been defined yet. So you aren't allowed to assume in your proof that $\lim_{x \to 0} e^x = e^0$; that would be circular.

Instead, you can find many questions on mathSE which go through the details of the proof. One of them is here.

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  • $\begingroup$ What if we have defined that this $exp$ is same as $e^{x}$ in our readings? Would you say I was allowed to do it then? $\endgroup$ – tenepolis Aug 11 '16 at 13:14
  • $\begingroup$ @tenepolis It is $99.9\%$ likely you are not allowed to assume it :/ It doesn't matter whether it's called $\exp(x)$ or $e^x$. The point is that this question is specifically about assuming the properties of the function in the first place. So you cannot assume them, or else it will be circular reasoning. In a math class you will have to get used to having to prove everything from "simpler" results, which most importantly means you can't assume as a result something you are trying to prove. So you can't just use any theorem or result you find in the readings. $\endgroup$ – 6005 Aug 11 '16 at 13:17

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