Using the $\varepsilon-\delta$ definition show that $\displaystyle \lim_{x \to a} x^n = a^n$
What I've done? Pretty much stuck on the step where one usually constructs delta:
$\displaystyle |x^n-a^n| < \varepsilon \implies \bigg|x-a\bigg|\bigg|\sum_{k=0}^{n-1}x^k a^{n-k-1}\bigg| <\varepsilon \implies |x-a| < \frac{\epsilon}{ \bigg| \displaystyle \sum_{k=0}^{n-1}x^k a^{n-k-1}\bigg|}$.
At this stage I cannot bound the denominator. Any suggestions are appreciated.
You want to have:
$$\delta < \frac{\epsilon}{\text{something}} < \frac{\epsilon}{|\sum_{k=0}^{n-1} x^k a^{n-k-1}|}$$
Then, $\text{something}$ must be greater than $|\sum x^k a^{n-k-1}|$. If we choose $\delta < 1$, then we would have for $|x - a| < \delta < 1$, $|x| < 1 + |a|$.
$$|\sum_{k=0}^{n-1} x^k a^{n-k-1}| \le \sum_{k=0}^{n-1}|x|^k |a|^{n-k-1} < \sum_{k=0}^{n-1} (1 + |a|)^k |a|^{n-k-1} < \sum_{k=0}^{n-1} (1+|a|)^k (1+|a|)^{n-k-1} \le \sum_{k=0}^{n-1} (1+|a|)^{n-1} = n(1 +|a|)^{n-1}$$
Thus, choosing:
$$\delta = \min\{1, \frac{\epsilon}{n(1+|a|)^{n-1}}\}$$
does the job.