Does there exist an epsilon-delta proof of the limit for all polynomials? I was wondering why there are different approaches to finding the existence of the limit for, say $\lim_{x \to a} x$, $\lim_{x \to a} x^2$, and $\lim_{x \to a} x^3$, if you can just cram lots of powers of $a$ in the denominator of $\delta$ instead. For example, to find
$$
\lim_{x \to a} x^n
$$
in the restricted case $a > 1, n \ge 1$, I tried setting
$$
\delta = (1 + \epsilon / a^n)^{1/n} - 1 > 1^{1/n} - 1 = 0
$$
Since $a > 1$, so if $a < x$, $x - a < \delta \implies x < \delta + a$ and $a > x \implies x < a + \delta$,
$$
0 < |x-a| < \delta \implies x < a + \delta
$$
Since $a > 1$,
$$
|x^n - a^n| < |(a + \delta)^n - a^n| < |(a + a\delta)^n - a^n| = |a^n((1 + \delta)^n - 1)|
$$
$$
|x^n - a^n| < |a^n((1 + (1 + \epsilon / a^n)^{1/n} - 1)^n - 1)| = |a^n((1 + \epsilon / a^n)^{1/n})^n - 1)|
$$
$$
|x^n - a^n| < |a^n(1 + \epsilon/a^n - 1)| = |a^n(\epsilon / a^n)| = |\epsilon|
$$
Since $\epsilon > 0$,
$$
|x^n - a^n| < \epsilon
$$
What I want to know is, is there a flaw in this proof? I feel like it can be extended pretty easily to $a < -1$, but is it much harder to find a suitable $\epsilon$ when $|a| < 1$? Or is the reason why nobody does this because the purpose of introducing epsilon-delta proofs is to formalize the definition of a limit, and so there's no point in actually calculating anything beyond $lim_{x \to a} x$?
 A: There is no flaw in your proof. However I agree that it is harder to extend this to the case $|a|<1$. 
Of course there are ways to get around it. First you can use 
$$\lim_{x \to a} f(x) g(x) = \lim_{x\to a} f(x) \lim_{x\to a} g(x)$$ 
and induction on $n$ to evaluate that. Or you can use the equality
$$x^n - a^n = (x-a) (x^{n-1} + x^{n-2} a + \cdots x a^{n-2} + a^{n-1}).$$
If you use this and triangle inequalities, we have 
$$|x^n-a^n| \le |x-a| \ \left( |x^{n-1}| + |x^{n-2}| |a| + \cdots + |x| |a^{n-2}| + |a^{n-1}| \right).$$
So for every $\epsilon>0$, first we decide to choose $\delta$ so that $\delta <1$. Then we have $|x| = |x-a + a| \le |x-a| + |a| < 1 + |a|$. Putting this into the inequality
$$\begin{split} 
&|x^{n-1}| + |x^{n-2}| |a| + \cdots + |x| |a^{n-2}| + |a^{n-1}|\\
&\le (|a|+1)^{n-1} + (|a|+1)^{n-2} |a| + \cdots (|a|+1) |a|^{n-2} + |a^{n-1}| \\
&=: C
\end{split}$$
The point is that this $C$ does not depends on $x$ (only on $a$ and $n$). Then if we further choose $\delta < \epsilon/C$, then 
$$|x^n - a^n | \le |x-a| C < \epsilon$$
whenever $|x-a| <\delta$. (That is, we choose $\delta = \min\{1, \epsilon/C\}$). 
