# Why is real exponentiation continuous in the base?

I know that real exponentiation is continuous in the exponent ($f(x)=a^x$ is continuous), but how do we know real exponentation is continuous in the base?

What I mean is, if $r$ is an arbitrary real number, why is the function $f(x):(0,\infty)\to\mathbb{R}$ given by $f(x)=x^r$ continuous? I define
$$x^r=\lim_{n\to\infty} x^{q_n}$$
where $(q_n)$ is any sequence of rationals converging to $r$.

I noticed that $\lim_{x\to 1}x^n=1$ for any integer $n$, by repeated use of the fact that the limit of a product of continuous functions is the product of the limit of each function. If $n$ is negative, the same thing follows by taking the limit of $1/x^{-n}$.

If $r$ is real, there exists integers $n,m$ such that $n\leq r\leq m$. By the Squeeze Lemma,
$$1=\lim_{x\to 1}x^n\leq\lim_{x\to 1}x^r\leq\lim_{x\to 1}x^m=1.$$
So I think $\lim_{x\to 1}x^r=1$ for any real number $r$. Can this be manipulated to show $f$ is continuous on all of $(0,\infty)$?

Please assume I don’t know any calculus or about $e$ or logarithms, only rudimentary results of limits in analysis. I’m trying to avoid anything like “since $x^r$ is differentiable, it is continuous, QED.” Thank you.

#### Solutions Collecting From Web of "Why is real exponentiation continuous in the base?"

If (by whatever definition of $a^b$ you have, provided it gives you the usual laws) you know $\lim_{x\to 1} x^b=1$ then $$\lim_{x\to a} x^b=\lim_{x\to 1} (ax)^b=\lim_{x\to 1} a^bx^b=a^b\lim_{x\to 1} x^b=a^b.$$