Continuity of $f(x) = x^n$ I want a $\epsilon-\delta$ proof for the continuity of $f(x)=x^n$ around a point $a$ on the domain of this function. 
Here's my sketch:
$$|f(x) - f(a)| = |x^n - a^n| = |x-a||x^{n-1} + x^{x-2}a  +\dotsb + xa^{n-2} + a^{n-1}|$$
For every pair $x^{n - j}a^{j-1} + x^{j-1}a^{n-j}$ we can write as 
\begin{multline}
x^{j-1}a^{j-1}\bigl((x^{n-2j +1} - a^{n-2j+1}) + 2a^{n-2j+1} \bigl)\\ = x^{j-1}a^{j-1}\bigl((x - a)(x^{n-2j} + x^{n-2j -1}a +\dotsb + xa^{n-2j-1} + a^{n-2j}) + 2a^{n-2j+1} \bigl)
\end{multline}
for every $j = 1,2,..., n$ 
Since $|x-a|<\delta$:
$$\bigl|x^{j-1}a^{j-1}\bigl((x - a)(x^{n-2j} + x^{n-2j -1}a +\dotsb + xa^{n-2j-1} + a^{n-2j}) + 2a^{n-2j+1} \bigl)| < |x^{j-1}a^{j-1}| \bigl( \delta (x^{n-2j} + x^{n-2j -1}a +\dotsb + xa^{n-2j-1} + a^{n-2j}) + |2a^{n-2j+1}| \bigl)$$ 
From now on I can't think of anything that would help me to conclude, I can't find any pattern that would help me summarize it, I would be glad if you help me.
Thanks.
 A: A few standard idioms (proofs left to you) will help.


*

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

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

*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, $|x| \geq \bigl||a| - 1\bigr|$.

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

*If $|x - a| \leq 1$, then the triangle inequality gives
$$
|x^{n-1} + x^{x-2}a  + \dots + xa^{n-2} + a^{n-1}| \leq n (1 + |a|)^{n-1}.
$$
A: Hint:
Establish as a lemma that the product of two continuous function is continuous, using
$$|f(x)g(x)-f(a)g(a)|=|f(x)g(x)-f(x)g(a)+f(x)g(a)-f(a)g(a)|\\<|f(x)|\,|g(x)-g(a)|+|f(x)-f(a)|\,|g(a)|.$$
