$\lim_{x\to a} x^{1/n}$

Question

How to find $\lim_{x\to b} x^{1\over n}$ $(n \in \mathbb{Z})$ without using that this function is continuous and without L'Hôpital's rule? I tried to approximate it like this: $\alpha$ > 0: Let x = b + $\alpha$, then $(b + \alpha)^{1\over n} \leq$ $b^\frac{1}{n}$ + $ \frac{\alpha}{n \cdot b^{1\over n-1}}$ But I don't know how to approximate when $\alpha \le 0$? Please, Help!

Answer 1

For $b>0:$

Use instead $u^n-v^n=(u-v)(u^{n-1}+u^{n-2}v+\cdots + v^{n-1})$, with $u=x^{1/n}$ and $v=b^{1/n}$. With $b>0$, at least, this will give you:

$$\left|x^{1/n}-b^{1/n}\right|=\frac{|x-b|}{x^{(n-1)/n}+x^{(n-2)/n}b^{1/n}+...+b^{(n-1)/n}}$$

If we bound $x>\frac{b}{2}$ then we get:

$$x^{(n-1)/n}+x^{(n-2)/n}b^{1/n}+\cdots+b^{(n-1)/n}>nb^{(n-1)/n}/2$$

and so:

$$\left|x^{1/n}-b^{1/n}\right| \leq \frac{2|x-b|}{nb^{(n-1)/n}}$$

Now, given $\epsilon>0$ pick a $\delta$ appropriately.