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## Homework Statement

Suppose f is continuous, f(1) = 5, and f(nx) = [f(x)]^n where n is any integer and x is any real number. Prove that f(x) = 5^x for all real x.

## Homework Equations

## The Attempt at a Solution

I've proved that f(x) = 5^x for rational x. Now I have to extend this to irrational x as well. Let x_n be a sequence of rational numbers converging to an irrational number z. By the continuity of f we have [tex] f(z) = \lim_{x_n \rightarrow z} f(x_n) = f(\lim_{x_n \rightarrow z} x_n) \Rightarrow f(z) = \lim_{x_n \rightarrow z} 5^{x_n} = 5^{\lim_{x_n \rightarrow z}} = 5^z [/tex].

My question is, how can I rewrite this proof to make use of the epsilon-delta definition of limit. Here is my attempt for that:

Since x_n converges to z and f is continuous, then [tex] |f(x_n) - f(z)| = |5^{x_n} - f(z)| < \epsilon [/tex] whenever [tex] 0 < |x_n - z| < \delta [/tex].

If you take f(z) = 5^z then |5^z||5^(x_n - z) - 1| < epsilon. 5^(x_n - z) has a limit of 1 and so the |5^(x_n - z) - 1| goes to 0, and so the quantity on the left hand side of the inequality can be made as small as we please, and so f(x) = 5^x for irrational x too.