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Suppose we know the definition of rational exponents of positive real numbers. If we define the exponential function $y=a^x\ (a>0)$ by extending rational exponents to real exponents by the following limiting process: $$a^x= \lim_{r_i(\in Q)\rightarrow x}a^{r_i}\ ,$$ then how can we prove the exponential function is continuous in $\mathbb{R}$?

I can't seem to write down anything meaningful...

Proton
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1 Answers1

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We need to show that $a^{x+\delta}=a^x+\epsilon$ with $\epsilon$ depending on $\delta$ and goes to $0$ as $\delta\to0$. Since $a^{x+\delta}-a^x=a^x(a^\delta-1)$ it is enough to focus on the second term.

For $a=1+h$, then by Bernoulli's inequality, $$a^\delta=(1+h)^\delta\le 1+h\delta$$ $$|a^\delta-1|\le|a-1||\delta|$$ So as $\delta\to0$, $a^\delta\to1$, and $a^{x+\delta}\to a^x$. Other cases such as $a<1$ can be handled similarly.

Chrystomath
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  • It seems Bernoulli's inequality need to be proved for irrational numbers in this situation. And I don't see how is the definition for irrational numbers used in this proof. – Proton Sep 22 '20 at 12:01
  • @JiangnanYu That follows from proving it for rational numbers, using that $(1+h)^\delta$ is increasing with $h,\delta$. – Chrystomath Sep 22 '20 at 12:17