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Let f(x)=a$^x$,x$\in \mathbb Q$ where a>0 and a$\in \mathbb R$.Is the function f uniformly continuous? If the function is uniformly continuous then given $\epsilon$>0 I have to find $\delta$>0 such that |a$^x$- a$^y$|<$ \epsilon$ whenever $|x-y|<\delta$. If I take x=$p\over q$ and y=$p_1\over q_1$ where q and $q_1$ are non zero,and proceed i'm not able to conclude anything...Plz help me

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  • $\begingroup$ i would suggest you to fix some $a$ and look for the behaviour... how does $a=2$ behave?? $f(1)=2;f(2)=4;f(3)=8$.....?? does this say something about uniform continuity? $\endgroup$ – user87543 Jan 9 '14 at 5:59
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Hints.

Before thinking about uniform continuity over $\mathbf{Q}$, think about it over $\mathbf{R}$ and then adapt your reasoning.

If I give you a certain fixed value, say $1$, is it possible to guarantee that $|a^y - a^x|$ will be less than 1 by taking $x$ and $y$ close enough? (It is understood that $x$ and $y$ can be chosen to be any numbers, as long as they are within a certain fixed distance of each other.)

For example, is there a value of $n$ such that if $y = x + 1/n$, you can be sure that $|a^y - a^x| \leq 1$?

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If you are familiar with differentiation, you can prove fairly easily that a sufficient condition for $f$ to be uniformly continuous is that $f'$ is bounded (and exists). You can use this criterion to show that $f$ is uniformly continuous on any bounded set.

Unfortunately, $f'$ is not bounded on $\mathbb{R}$. While it is not the case that any function $f$ with $f'$ unbounded is not uniformly continuous, the fact that $f'$ diverges is a strong hint that $f$ is not uniformly continuous, and tells you where to look for the counterexample. What can you tell about $f(x)-f(y)$ if $y =x+\delta$ and $x$ is very large (say, $x \to \pm \infty$, while $\delta$ remains fixed)?

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