Proving that $\lim_{x\rightarrow a} \sqrt[n]x = \sqrt[n]a$ using $\epsilon$ and $\delta$

Question

How do I show that $\lim\limits_{x\rightarrow a} \sqrt[n]x = \sqrt[n]a$ using an $\epsilon$-$\delta$ proof, where $n$ is a positive integer?

I know the definition of a limit, so I want to show that $|\sqrt[n]x - \sqrt[n]a| < \epsilon$ using the fact that $|x-a|<\delta$ for some $\delta > 0$. How do I approach this? I've been stuck on this for a while.

Thanks in advance.

Answer 1

Hint: Use the identity $$p^n-q^n=(p-q)(p^{n-1}+p^{n-2}q+\cdots+pq^{n-2}+q^{n-1})$$ then, $$\sqrt[n]{x}-\sqrt[n]{a}=\frac{x-a}{P}$$ where $P$ is ...