Continuity of exponential functions (epsilon-delta)

[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?

 

[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?