Continuity of $f(x) = x^n$

Solution 1:

A few standard idioms (proofs left to you) will help.

1. You may as well assume $\delta \leq 1$. (If $\delta > 0$ is arbitrary, you can replace it with $\min(\delta, 1)$.)

2. If $x$ and $a$ are real numbers, then $$ |x - a| < \delta\quad\text{if and only if}\quad a - \delta < x < a + \delta. $$

3. Consequently, if $|x - a| \leq 1$, then $|x| \leq 1 + |a|$.

   You don't need it here, but there's also a useful lower bound: If $|x - a| \leq 1$, then $|x| \geq 1 - |a|$ and $|x| \geq |a| - 1$. Succinctly, $|x| \geq \bigl||a| - 1\bigr|$.

4. If $j$ and $m$ are integers such that $0 \leq j \leq m$, and if $|x - a| \leq 1$, then $$ |x^{j} a^{m-j}| = |x|^{j}\, |a|^{m-j} \leq (1 + |a|)^{m}. $$

5. If $|x - a| \leq 1$, then the triangle inequality gives $$ |x^{n-1} + x^{x-2}a + \dots + xa^{n-2} + a^{n-1}| \leq n (1 + |a|)^{n-1}. $$