Proving $x^n$ is continuous for all $n$

Question

I'm trying to prove that $f(x) = x^n$ is continuous for all $n$. One option is the Binomial theorem, but it seems slightly messy, so I've opted for induction. Here is my attempt.

Let $a \in \mathbb{R}$ be given and induct on $n$. If $n=1$, given $\epsilon = 0$, set $\delta = \epsilon$. Then $|x-a| < \delta$ implies $|f(x) - f(a)| = |x - a| < \delta = \epsilon$, so $f(x)$ is continuous at $a$ and thus on $\mathbb{R}$.

If $n = 2$, we need to show that $f(x) = x^2$ is continuous at $a$. This is where I am having trouble, because the scratchwork falls apart. If $|x-a| < \delta$, then $$|f(x) - f(a)| = |x^2 - a^2| = |(x+a)(x-a)| = |x+a||x-a| < \delta |x+a|.$$ $\delta$ can depend on $a$, but not on $x$, so I don't know what kind of additional restrictions I can place on $\delta$.

The induction is straightforward from here, but only if I don't use the definition, but rather the fact that $x^k$ is continuous by assumption, as is $x$, and the product of two continuous functions is continuous. I'd prefer to do this from the definition, however, as this seems to trivialize the result. I could have proved that $x^2$ is continuous just by using the $n=1$ case.

Help would be appreciated.

Does this answer your question? How can I prove that a polynomial with degree $n$ is continuous everywhere in $\mathbb{R}$ using definitions? — Eminem, Jan 27 '21 at 07:56
How about proving that it is differentiable based on my answer here https://math.stackexchange.com/questions/4001006/elementary-proof-of-power-rule-for-differentiation-fx-rxr-1-for-fx-x/4001168?noredirect=1#comment8257908_4001168 and then you can conclude continuity. — B E I R U T, Jan 27 '21 at 07:58
Does that include $n<0$? It's still true, though maybe less obvious. — Jean-Claude Arbaut, Jan 27 '21 at 08:14

score 1 · Answer 1 · answered Jan 27 '21 at 08:28

Hint

Continuity at $0$ is simple as for $\vert x \vert \le 1$ your have $\vert x^n \vert \le \vert x \vert$.

And following your idea of induction to prove continuity at $a \neq 0$

$$\vert x^{n+1} - a^{n+1} \vert = \vert (x-a)(x^n + x^{n-1}a + \dots ++ a^{n}) \vert \le \vert x - a \vert (n+1)2^n \vert a \vert^n$$ for $ \vert x \vert \le 2 \vert a \vert$.

You can apply that to the case $n \ge 2$. Based on the inequality, proof of continuity by induction is straightforward.

Jean-Claude Arbaut · Answer 2 · 2021-01-27T08:42:40.217

Let $f$ and $g$ be two functions on some open domain $D$, continuous at $x_0\in D$, and let $h=fg$. Let's prove $h$ is continuous at $x_0$.

For any $\varepsilon>0$, let $\varepsilon'=\dfrac{\min(\sqrt{\varepsilon},\varepsilon)}{3\max(1,|f(x_0)|,|g(x_0)|)}$ (to simplify later).

Then since $f$ is continuous at $x_0$ there is a $\delta_1>0$ such that $|x-x_0|\le\delta_1\implies |f(x)-f(x_0)|\le\varepsilon'$.

Likewise, since $g$ is continuous at $x_0$ there is a $\delta_2>0$ such that $|x-x_0|\le\delta_2\implies |g(x)-g(x_0)|\le\varepsilon'$.

Now let $\delta=\min(\delta_1,\delta_2)$, we have, for $|x-x_0|\le\delta$,

$$|f(x)g(x)-f(x_0)g(x_0)|=|(f(x)-f(x_0))(g(x)-g(x_0))\\+f(x_0)(g(x)-g(x_0))+g(x_0)(f(x)-f(x_0))|\le \varepsilon'^2+|f(x_0)|\varepsilon'+|g(x_0)|\varepsilon'\le\varepsilon$$

Hence $fg$ is continuous at $x_0$.

Now, since $x\to1$ and $x\to x$ are trivially continuous everywhere, you can prove by induction that $x\to x^n$ is continuous everywhere, for all $n\in\Bbb N$.

Proving $x^n$ is continuous for all $n$

2 Answers2