# Use the Definition of a limit to prove $\lim 3^{1/n}=1$

I know the definition of finite limit that $\lim a_n =L$ if $\forall \epsilon > 0$, $\exists n_{\epsilon} \in \mathbb{N}$ such that $\forall n> n_{\epsilon}$, $|a_n - L| < \epsilon$. So I can start with

$$\left| 3^{1/n} - 1 \right| < \epsilon$$

and then I'm pretty stuck. I can try to exponentiate the first term $$\left| e^\frac{\ln 3}{n} - 1 \right| < \epsilon,$$ but I don't see how it helps...

• For each epsilon you can find N such that the nth (n> N) power of numbers larger than $1+\epsilon$ is larger than three and the nth (n> N) power of numbers smaller than $1-\epsilon$ is smaller than 3. Prove that and then finish – Bananach Aug 13 '16 at 17:18

## 1 Answer

HINT:

Recall ( SEE THIS ANSWER )that for $x<1$, the exponential function satisfies the inequalities

$$1+x\le e^x\le \frac{1}{1-x}\tag 1$$

Can you finish now?

SPOILER ALERT: Scroll over the highlighted area to reveal the solution.

Using $(1)$ we can write for $n\ge 2$ \begin{align}\left|3^{1/n}-1\right|&=\left|e^{\frac1n \log(3)}-1\right|\\\\ & \le \frac{\frac1n \log(3)}{1-\frac1n \log(3)}\\\\&=\frac{\log(3)}{n-\log(3)}\\\\ &<\epsilon \end{align} whenever $n>\max\left(2,\log(3)+\frac{\log(3)}{\epsilon}\right)$. And we are done!

• Thank you! The assertion $e^x \le 1/(1-x)$ is trivial? I can see that $1+x \le e^x$ because $1+x$ is just first order approx. of $e^x$ at $0$, but the second inequality isn't that obvious for me. – J.Doe Aug 13 '16 at 17:24
• You're welcome. My pleasure. I've added a link to an answer in which I showed both inequalities using only the limit definition of the exponential function and Bernoulli's Inequality. – Mark Viola Aug 13 '16 at 17:26