# Limit of a function using epsilon delta

calculuslimits

I'm trying to prove that, for some constant $$x_0$$

$$\lim_{n \to \infty}\frac{1}{(x-x_0)^{\frac 1 n}} \to 1$$

where I verified the result above with Wolfram Alpha.

I am struggling in the epsilon delta though. Let $$\varepsilon > 0$$ be given. The problem is that $$n \to \infty$$ so I'm not sure if I should be doing $$|n – \infty| < \delta \implies \left|\frac{1}{(x-x_0)^{\frac 1 n}} – 1\right| < \varepsilon$$ or how that $$|n – \infty|$$ is defined. How should I proceed?

This isn't an $$\epsilon - \delta$$ proof but rather a standard sequence convergence proof. Further, the limit exists for all $$x>x_0$$ so I'm not sure why that is part of the question. What you are really trying to prove is that for any positive number $$a$$ we have

$$\lim_{n\to\infty}a^{1/n}=1$$

There are many ways to prove this but I prefer this way: We have that

$$a^{1/n}=e^{\ln(a^{1/n})}=e^{\frac{\ln(a)}{n}}$$

Since $$\ln(a)$$ is a real number and $$e^x$$ is continuous at $$x=0$$ we have

$$\lim_{n\to\infty}a^{1/n}=\lim_{n\to\infty}e^{\frac{\ln(a)}{n}}=e^{\lim_{n\to\infty}\frac{\ln(a)}{n}}=e^0=1$$

Of course, this proof might not suffice for you depending on what you have already proved but its a good start nonetheless.