How do I prove, using the definition, that the nth root is a continuous function?

Question

Stuff I've tried: \begin{align} |x^{\frac 1 n}-a^{\frac 1 n}| &< \epsilon \\ |(x^{\frac 1 n}-a^{\frac 1 n})^n| &< \epsilon ^n \\ \left|\sum_{i=0}^{n}C(n,i)x^{\frac in}(-a)^{1-\frac i n}\right| &< \epsilon ^n \end{align}

(it looks like a one but the power on $x$ and $a$ it's actually $\frac i n$)

if we take $\delta \leq 1$

$$ |x| < |a| + 1 \\ \left|\sum_{i=0}^{n}C(n,i)x^{\frac in}(-a)^{1-\frac i n}\right| \leq 2^n(|a| + 1) $$

I would love to be able to say that $\delta = \min \left\{1, \frac \epsilon{2^n(|a|+1)} \right\}$ but this doesn't seem to be the case. I also tried using the $$x^n - y^n = (x-y)(x^{n-1} + ... + y^{n-1})$$ formula but it didn't help much either.

The way I used this formula for $n=3$ for example was: $$|\sqrt[3]x - \sqrt[3]a| < \epsilon \\ |x-a| < \epsilon ( \sqrt[3]{x^2} + \sqrt[3]xa + \sqrt[3]{a^2}) $$ if we use the same trick $\delta \leq 1$ $$( \sqrt[3]{x^2} + \sqrt[3]xa + \sqrt[3]{a^2}) < 3\sqrt[3]{(|a|+1)^2}$$ but even if we set $\delta = \min \left\{1, 3\epsilon \sqrt[3]{(|a|+1)^2}\right\}$ it can't be shown that $|x-a|< \delta$ implies $|\sqrt[3]x - \sqrt[3]a| < \epsilon$

Prove continuity for cubic root using epsilon-delta

Answer 1

Since you've given a general $n$th root, I'm asssuming $a\ge 0$. If $a = 0$, continuity follows from the fact that for every $\epsilon > 0$, setting $\delta = \epsilon^n$ forces $|x^{1/n}| < \epsilon$ whenever $|x| < \delta$. Now suppose $a > 0$. Let $\epsilon > 0$. Let $\epsilon'$ be such that $0 < \epsilon' < \min\{a^{1/n}, \epsilon\}$. Set $\delta = \min\{(a^{1/n} + \epsilon')^n - a, a - (a^{1/n} - \epsilon')^n\}$. Then $$a - \delta > a - [a - (a^{1/n} - \epsilon')^n] = (a^{1/n} - \epsilon')^n$$ and $$a + \delta < a + [(a^{1/n} + \epsilon')^n - a] = (a^{1/n} + \epsilon')^n.$$ So for all $x$, $|x - a| < \delta$ implies $(a^{1/n} - \epsilon')^n < a - \delta < x < a + \delta < (a^{1/n} + \epsilon')^n$, which implies $a^{1/n} - \epsilon' < x^{1/n} < a^{1/n} + \epsilon'$, i.e., $|x^{1/n} - a^{1/n}| < \epsilon'\le\epsilon$. Since $\epsilon$ was arbitrary, the result follows.

Answer 2

With some accessible work, you can also show the inequality $$ \left|x^{1/n}-y^{1/n}\right|\le|x-y|^{1/n} $$ showing Hölder continuity and uniform continuity on $[0,\infty)$.

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The proof starts with considering $0\le y\le x$, so that $x^{1/n}=y^{1/n}+h$, $h\ge0$ and uses only the binomial theorem, $$ x-y=(y^{1/n}+h)^n-y=((y^{1/n})^n+...+h^n)-y\ge h^n \\ \implies\sqrt[n]{x-y}\ge x^{1/n}-y^{1/n} $$

Answer 3

Hints: Let's have some ideas: for $n = 2$ we have:$$|\sqrt{x}-\sqrt{a}| = \frac{|x-a|}{\sqrt{x}+\sqrt{a}} \leq \frac{|x-a|}{\sqrt{a}},$$

and $\delta < \epsilon\sqrt{a}$ will do. For $n = 3$ we have: $$|\sqrt[3]{x}-\sqrt[3]{a}| = \frac{|x-a|}{\sqrt[3]{x^2} + \sqrt[3]{xa} + \sqrt[3]{a^2}} \leq \frac{|x-a|}{\sqrt[3]{a}^2},$$ if $a > 0$, so $\delta = \min\{a, \epsilon\sqrt[3]{a}^2\}$ solves it. Do it the same way for $a < 0$ ($a = 0$ is easy).

How it would be for $n = 4$? For $n$ even it is easier, we don't have to break anything into cases.

Notice that I'm using all the way that: $$\begin{align}x-y &= (x^{1/n})^n-(y^{1/n})^n \\ &= (x^{1/n}-y^{1/n})(x^{(n-1)/n}+x^{(n-2)/n}y^{1/n}+\ldots+x^{1/n}y^{(n-2)/n}+y^{(n-1)/n}).\end{align}$$