1. ## limit proof

I want to check whether I've proved the following limit question correctly. Because i don't feel so.
$
\lim_{x \to \infty} 3^{\frac{1}{x}}$

attempt:

let $\epsilon >0$ and $a>0$

$x > a \implies \frac{1}{x}<\frac{1}{a} \implies 3^{\frac{1}{x}}-1<3^{\frac{1}{a}}-1$

so define $a=\frac{1}{\log_3(\epsilon+1)}$

then $x>a \implies 3^{\frac{1}{x}}-1<\epsilon \implies |3^{\frac{1}{x}}-1|<\epsilon$

so can anyone check my work on this

2. It looks fine.

3. yet i can prove $\lim_{x \to \infty} 3^{\frac{1}{x}} = 0$ in the same way?

yet i can prove $\lim_{x \to \infty} 3^{\frac{1}{x}} = 0$ in the same way?
I doubt it...

You can prove it in other way if you don't want to use the formal definition...

$\displaystyle \lim_{x \to \infty}3^{\frac{1}{x}} = \lim_{x \to \infty}e^{\ln{\left(3^{\frac{1}{x}}\right)}}$

$\displaystyle = \lim_{x \to \infty}e^{\frac{1}{x}\ln{3}}$

$\displaystyle = e^{\lim_{x \to \infty}\frac{\ln{3}}{x}}$

$\displaystyle = e^0$

$\displaystyle = 1$.

5. $log_3(\epsilon +1)>0$ is important. For small $\epsilon$, $log_3(\epsilon )<0$ which is where the argument fails for 0.

6. Also, in your argument you need to look at $3^\frac{1}{x}-1>-\epsilon$ more carefully. It works out fine here, but this is where the argument might fall apart if you try to prove the limit is equal to something larger than 1.

The important thing here is that $1-\epsilon$ is always less than 1, so that you always get a negative number when you take the log (and 1 is the largest number that has this property for all $\epsilon$).