Is this the wrong way to find $\delta$? Find $\delta$ such that $$|x^n - a^n| < \epsilon $$ whenever $$|x-a|<\delta$$
My thought is factoring $|x^n-a^n|$ so we have $|(x-a)|\cdot|x^{n-1}+x^{n-2}a+x^{n-3}a^2+...+xa^{n-2}+a^{n-1}|$
In order to make it smaller than $\epsilon$, I let $|x-a|<1$, so I have $$(a-1)^k<x^k<(a+1)^k$$
Thus, 
$$|(x-a)|\cdot |(a-1)^{n-1}+(a-1)^{n-2}a+..+a^{n-1}| <|(x-a)|\cdot|x^{n-1}+x^{n-2}a+...+xa^{n-2}+a^{n-1}| \hspace{0.5cm} (1)$$
so I choose $$\delta=\min \left(1,\frac{\epsilon}{|(a-1)^{n-1}+(a-1)^{n-2}a+..+a^{n-1}|}\right)$$
However, it strikes me that it might not be true, because inequality (1) might not hold when $a$ is negative.
 A: You're right that your inequality might not hold for lesser $x$, since the function $x\mapsto x^{n-1}+x^{n-2}a+x^{n-3}a^2+\ldots+a^{n-1}$ is only guarunteed to be increasing when $a$ and $x$ are positive. A much more important error you made is that you should be working with $(a+1)$ not $(a-1)$. In particular, your goal is to ensure that $|x^n-a^n|<\varepsilon$. Your reasoning at the moment is:

Since $|x-a|\cdot |k^{n-1}+k^{n-2}a\ldots+a^{n-1}|<|x-a|\cdot |x^{n-1}+x^{n-2}a+\ldots +a^{n-1}|$ in the desired interval for some $k$, if we bound the former by $\varepsilon$, then the latter will also be bounded by $\varepsilon$.

This isn't good, because it's trying to say that if $A<B$ and $A<C$, then $B<C$ - which you should see the issue with. What you want to do is to say:

Since $|x-a|\cdot |x^{n-1}+x^{n-2}a+\ldots +a^{n-1}|<|x-a|\cdot |k^{n-1}+k^{n-2}a\ldots+a^{n-1}|$ in the desired interval for some $k$, if we bound the latter by $\varepsilon$, then the former will also be bounded by $\varepsilon$.

which holds if you let $k=(a+1)$ and then $\delta=\min(1,\frac{\varepsilon}{(a+1)^{n-1}+(a+1)^{n-2}a+\ldots+a^{n-1}})$. This, of course, only works when $x\geq 0$, but the argument for negative $x$ is fairly obvious from there (since $x^n$ is either an even or odd function)
