# How do I prove this?

1. Feb 26, 2005

### semidevil

find the limit

$$(3n)^{\frac{1}{2n}}$$ as n goes to infinity

just by eyeballing, I know that as the nget sufficiently large, the limit of the exponent will get to 0. and 3n to the 0 will be one....

im' sure on the exam, this wont work, and I dont know how to give a precise proof of it. this is the chapter on subsequences and Bolzano-wiestress theorem, so I assume I need to use one of these concepts...but I dont know how....

any tips?

Last edited: Feb 26, 2005
2. Feb 26, 2005

### AKG

$$(3n)^{\frac{1}{2n}} = (3n)^{\frac{1}{3n}\times \frac{3}{2}} = \left ((3n)^{\frac{1}{3n}}\right )^{1.5} = (x^{1/x})^{1.5}$$

$$\lim _{x \to \infty} x^{1/x} = \lim _{x \to \infty} \exp (\ln x^{1/x}) = \lim _{x \to \infty} \exp \left (\frac{\ln x}{x}\right ) = \lim _{x \to \infty} \exp \left (\frac{1/x}{1}\right ) = \lim _{x \to \infty} \exp (0) = 1$$

$$\lim _{n \to \infty} (3n)^{\frac{1}{2n}} = 1^{1.5} = 1$$

If you wrote the solution out exactly like this, you'd probably lose some marks, but the idea is right. You can figure out on your own how to write it out properly, I just did it this way to save time.

Last edited: Feb 26, 2005
3. Feb 26, 2005

### semidevil

thanx....great..but where did the

$$(x^{1/x})^{1.5}$$

come from?

4. Feb 26, 2005

### AKG

x = 3n......

5. Feb 27, 2005

### dextercioby

And one more thing:It's the theorem of Bolzano-WEIERSTRASS (sic)...

Daniel;

6. Mar 24, 2005

### DH

ok, but i think that i got another way
$$set y = \lim _{n \to \infty} (3n)^{\frac{1}{2n}}$$
we take = $$lny = ln( \lim _{n \to \infty} (3n)^{\frac{1}{2n}})$$
Since $$ln( \lim _{n \to \infty} (3n)^{\frac{1}{2n}}) is continuous function$$
--> $$lny = \lim _{n \to \infty}[ln (3n)^{\frac{1}{2n}}]$$
$$lny = \lim _{n \to \infty}[{\frac{ln3n}{2n}}]$$
$$infty/infty$$ -> use L'Hopital
$$lny = \lim _{n \to \infty}[{\frac{1}{2n}}]$$
--> lny = 0 -> y =1

7. Mar 24, 2005

### dextercioby

The method is correct and so is the result.The trick with the commuting between "ln" (or "exp") and the limit is very useful & u proved it.

Daniel.

8. Mar 29, 2005

### Data

$$\lim_{x \rightarrow \infty} e^{\frac{\ln{x}}{x}} = e^{\lim_{x \rightarrow \infty} \frac{\ln{x}}{x}}$$

and

$$\lim_{x \rightarrow \infty} \frac{\ln{x}}{x} = \lim_{x \rightarrow \infty} \frac{\frac{1}{x}}{1} = 0$$

by l'Hopital.