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Show the following: Let a>0 be a real number, then f:R->R defined by [itex]f(x) = a^x[/itex] is continuous.

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First we have the following definitions of continuity:

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Let X be a subset ofR, let f:X->R be a function, and let [itex]x_0[/itex] be an element of X. Then the following three statements are logically equivalent:

(a) f is continuous at [itex]x_0[/itex]

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

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

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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 [itex]x_0[/itex]). What I tried was the following, but it does not look pretty:

Let [itex]x_0[/itex] be some point in R. Our function is increasing, so I said that if we look at [itex]x_0 - \delta[/itex] and [itex]x_0 + \delta[/itex] then we want to show the following: [itex]|max(a^{x_0 - \delta},a^{x_0 + \delta}) - a^{x_0}| < \epsilon[/itex] I then factored out the [itex]a^{x_0}[/itex] and started moving things around to get delta in terms of a,epsilon, and [itex]x_0[/itex], 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?

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# Homework Help: Continuity of exponential functions (epsilon-delta)

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