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Here is what I need to prove: Let $a>0$ be a positive real number. Then the function $f: \Bbb R\to \Bbb R$ defined by $f(x):=a^x$ is continuous.

We are not supposed to use logarithms. Some of the hints given are:

  1. For $a>0$, limit as $n$ goes to infinity of $x^{1/n}$ is $1$.
  2. Try to use the squeeze theorem.
ForgotALot
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Taylor
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1 Answers1

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It is sufficient to prove that $\lim_{x \to 0} a^x = 1$ since $|a^y - a^x| = |a^x||a^{y-x}-1|.$

The construction of the exponential function generally proceeds by defining it first for rational exponents. Extension to irrational exponents $x$ follows by taking $a^x = \lim_{n \to \infty}a^{r_n}$ where the rational sequence converges monotonically to $x$. All that is needed here is the property that $a^x$ is monotone increasing with respect to $x$ if $a > 1$ and monotone decreasing if $0 < a < 1$.

For any number $x \in (-1,1)$ there exists an integer $N$ such that $-1/N < x < 1/N$.

Hence, if $a > 1$

$$a^{-1/N} < a^x < a^{1/N},$$

and the squeeze theorem shows that $a^x \to 1$.

If $0 < a < 1$, then the same argument applies to $1/a$.

RRL
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