# Continuity of exponential functions (epsilon-delta)

1. Nov 27, 2006

### mattmns

Here is the question:
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Show the following: Let a>0 be a real number, then f:R->R defined by $f(x) = a^x$ is continuous.
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First we have the following definitions of continuity:
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Let X be a subset of R, let f:X->R be a function, and let $x_0$ be an element of X. Then the following three statements are logically equivalent:

(a) f is continuous at $x_0$

(b) For every sequence $(a_{n})_{n=0}^{\infty}$ consisting of elements of X with $\lim_{n\rightarrow \infty}a_{n} = x_0$, we have $\lim_{n\rightarrow \infty}f(a_n) = f(x_0)$

(c) For every $\epsilon > 0$ there exists a $\delta > 0$ such that $|f(x) - f(x_0)| < \epsilon$ for all x in X with $|x - x_0| < \delta$

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I thought the best way to do this would be to use the epsilon-delta definition, since I really could not think of anything with the sequences (contradiction, maybe?).

So now the question is, what do we make delta. Clearly delta must depend on epsilon and on a (and I think on $x_0$). What I tried was the following, but it does not look pretty:

Let $x_0$ be some point in R. Our function is increasing, so I said that if we look at $x_0 - \delta$ and $x_0 + \delta$ then we want to show the following: $|max(a^{x_0 - \delta},a^{x_0 + \delta}) - a^{x_0}| < \epsilon$ I then factored out the $a^{x_0}$ and started moving things around to get delta in terms of a,epsilon, and $x_0$, but it did not really work out, and I am not sure where to go from here.

Any ideas? Thanks!

edit... I think I remember hearing something about showing that it is continuous for some explicit point (like x = 0) and then reducing every other case (meaning every other value x can take on) to the same case of x = 0, thus showing continuity. Would this be sufficient (I think so), and would that be a good way to solve this exercise?

Last edited: Nov 28, 2006
2. Nov 28, 2006

### StatusX

Start by showing a^x is continuous as a function from the rationals to the reals. How is a^x defined when x is irrational?