Proving of limits 
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*Let $a>0$ and $n$ be a positive integer. Prove that $$\lim_{x\to a} x^{1/n} = a^{1/n}$$


For this question, I'm not entirely sure whether I should proceed with the $\epsilon$-$\delta$ definition of a limit, or simply prove the power law for limits, since the question did not specify the need to use the precise definition of a limit in this proof. I'm thinking that I could try to prove that the function is continuous at $x=a$, but proving continuity requires $\lim_{x\to a} x^{1/n} = a^{1/n}$. Could someone hint me in the right direction for this?


*Let $a$ be a real number. Prove that 
(a) $lim_{x\to a^-} \frac{1}{(x-a)^3} = -\infty$ and
(b)$lim_{x\to a^+} \frac{1}{(x-a)^3} = \infty$
My attempt at the second question is as follows:
(a) We want to prove $\forall N<0$, $\exists \delta >0$ such that if $a - \delta < x < a$,  then $\frac{1}{(x-a)^3} < N$. $$\frac{1}{(x-a)^3}<N$$ $$\frac{1}{N} > (x-a)^3$$ so  $x-a > \sqrt[3]{\frac{1}{N}}$, i.e. $x > \sqrt[3]{\frac{1}{N}}$, so we choose $\delta = -\sqrt[3]{\frac{1}{N}}$. Note that $\sqrt[3]{\frac{1}{N}} + a < x < a$, so $\frac{1}{(x-a)^3} < N$ (Q.E.D)
(b) Similarly, we want to prove that $\forall M > 0$, $\exists \delta >0$ such that if $a<x<a+\delta$ then $\frac{1}{(x-a)^3} > M$. $$\frac{1}{M} > (x-a)^3$$ $$x < \sqrt[3]{\frac{1}{M}} + a$$, so we choose $\delta = \sqrt[3]{\frac{1}{M}}$. Note that $a<x<a+ \sqrt[3]{\frac{1}{M}}$, so $\frac{1}{(x-a)^3} > M$ (Q.E.D)
I feel like this proof I produced is missing something necessary to conclude the respective limits, or it could be completely wrong, so any criticism for the proof would be appreciated!
Thanks in advance!
 A: For question 1, use the following geometric series identity:
$$\frac{p^n - q^n}{p - q} = p^{n-1} + p^{n - 2} q + p^{n - 3} q^2 + \ldots + q^{n - 1}$$
with $p = x^{\frac{1}{n}}$ and $q = a^{\frac{1}{n}}$.
As for question 2, you're doing fine. There is what I think is an inconsequential typo under your first two inequalities, where you say $x > \sqrt[3]{\frac{1}{N}}$, where I think you may mean $x > a + \sqrt[3]{\frac{1}{N}}$, but it doesn't cause any problems down the road.
It's hard for me to gauge exactly what is bothering you; I can only guess. However, what slightly bothers me is that the proof is presented slightly backwards. It's presented in the order of your working out, rather than in logical order.
You're supposed to be able to take a number $N < 0$, and instantly produce a $\delta > 0$ with the given property. So, your proof should probably take the form:

Suppose $N < 0$, and let $\delta = -\sqrt[3]{\frac{1}{N}} > 0$. Then $a - \delta < x < a \implies \ldots \implies \frac{1}{(x - a)^3} < N.$

The working you've written is valuable working behind the scenes, but it's nicer to present the proof in the proper order.
EDIT: In response to your comment, that's not too hard to do, and I think it is worth covering. Anything you feel is shaky is worth delving into. A proof is supposed to be convincing, and if it can't even convince the person who wrote it, then it's not working.
To prove this, suppose that $a + \sqrt[3]{\frac{1}{N}} < x < a$. Then
$$x - a > \sqrt[3]{\frac{1}{N}}.$$
Then, because the function $x \mapsto x^3$ is increasing, we have
$$(x - a)^3 > \left(\sqrt[3]{\frac{1}{N}}\right)^3 = \frac{1}{N}.$$
Now, $x < a$, so $x - a < 0$. For negative numbers, the map $x \mapsto \frac{1}{x}$ is decreasing, so
$$\frac{1}{(x - a)^3} < N,$$
which is what we wanted to prove.
A: For $(1)$ you can use this lemma:

Let $x > y \ge 0$ and $n \in \mathbb{N}$. Then $$x^{1/n} - y^{1/n} \le (x-y)^{1/n}$$

Proof:
Let $t = \left(\frac{y}x\right)^{1/n} \in [0,1\rangle$. Then $$(1-t)^n \le 1-t \le 1-t^n$$
so $1-t \le (1-t^n)^{1/n}$. Multiplying with $x^{1/n}$ gives the desired result.
Now you should be able to finish with
$$\left|a^{1/n} - x^{1/n}\right| \le |a-x|^{1/n}$$ 
Your attempt at $(2)$ looks good.
