Proving the Limit of (3n)^{\frac{1}{2n}} as n Goes to Infinity

[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 won't work, and I don't 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 don't know how...

any tips?

 

[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.

 

[semidevil]

(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.

thanx...great..but where did the

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

come from?

 

[AKG]

x = 3n...

 

[dextercioby]

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

Daniel;

 

[DH]

(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.

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

 

[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.

 

[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.