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Consider the function $e^x$ on the reals. I want to show that $e^x$ is continuous at the any point $t \in \mathbb{R}$. Is the following argument valid?

(1) Let $\{t_n\}$ be an arbitrary sequence of reals s.t. $t_n \to t$ and $t_n \ne t$.

(2) Then $\lim\limits_{t_n \to t}$ $e^{t_n} = \lim\limits_{t_n \to t}1 + \frac{{t_n}^2}{2} + \frac{{t_n}^3}{6} + \ldots = 1 + \frac{{t}^2}{2} + \frac{{t}^3}{6} + \ldots = e^t$

So that since $\{t_n\}$ is arbitrary it follows $e^x$ is continuous at $t$.

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You have to tell how do you define the exponential function, because the proof will depend greatly on the definition you are using. For example, if we define the exponential functions as a solution of the following Cauchy problem: $y'=y$ and $y(0)=1$, then we do not really have to prove anything... –  Godot Oct 29 '12 at 22:37
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2 Answers

up vote 2 down vote accepted

Your proof is not satisfactory, unless you use some theory about infinite sums of functions and the continuity of them. Namely:

THEOREM Let $\{f_k\}$ be a sequence of continuous functions. Set $$F_n=\lim_{n\to\infty }\sum_{k=0}^n f_k$$

If $F_n$ converges to a function $F$ uniformly, then $F$ is continuous.

In your case, you have $$f_k=\frac{x^k}{k!}$$

and convergence is indeed uniform, so $e^x$ is continuous, but this seems a little too much for what you want to prove.

Instead, you can consider that $e^x$ is differentiable over all $\Bbb R$; or that $$e^x=\int_0^x e^t dt +1$$ or consider that $$e^{x+h}-e^x=e^x(e^h-1)$$ so it suffices to show $e^x$ is continuous at $x=1$. Another approach would be to show that $\log x$ is continuous.

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A very elementary and interesting exposition, related with this question, can be found in the book "Calculus, Early Trascendentals" by James Stewart, 4th Edition, Section 1.5 "Exponential Functions", pages 56 and 57. It is not a demonstration at all, it begins by defining 2^x for integers, then for rational numbers, finally for irrational numbers, and by construction it is at least plausible that 2^x must be continuous. May be not what you need, yet I believe it is fun to read.

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