# Finding Limits Using L'Hopital's Rule

• September 3rd 2012, 09:12 AM
Preston019
Finding Limits Using L'Hopital's Rule
Okay I'm trying to find this limit:

lim (ln(x))1/x as x approaches infinity using L'Hopital's rule.

This is what I've done so far:

y = (ln(x))1/x

ln(y) = (ln(ln(x)))/x

lim ln(y) = lim (ln(ln(x)))/x (Both as x approaches infinity)

(using L'H) = lim 1/(xln(x)) (as x approaches infinity)

That is what I have so far and I do not know where to go from there.
Also, sorry for the bad use of writing the equations and what not; I'm not sure how to write them on here as they should actually look, if that makes sense.
• September 3rd 2012, 09:25 AM
SworD
Re: Finding Limits Using L'Hopital's Rule
Both x and ln(x) approach infinity as x approaches infinity. Therefore if you are dividing by them, the reciprocal has to approach 0.

Edit: typo. The reciprocals, in this case 1/x and 1/ln(x) both have to approach 0, so 1/(x*ln(x)) approaches 0 as well.
• September 3rd 2012, 10:58 AM
MaxJasper
Re: Finding Limits Using L'Hopital's Rule
$\lim_{x\to \infty } \, \sqrt[x]{\log (x)} = 1$
• September 3rd 2012, 11:31 AM
Preston019
Re: Finding Limits Using L'Hopital's Rule
Quote:

Originally Posted by SworD
Both x and ln(x) approach infinity as x approaches infinity. Therefore if you are dividing by them, the reciprocal has to approach 0.

Edit: typo. The reciprocals, in this case 1/x and 1/ln(x) both have to approach 0, so 1/(x*ln(x)) approaches 0 as well.

True. But that's not the answer. We're supposed to find lim y as x approaches infinity. So far, all I have is what lim ln(y) as x approaches infinity equals. I'm supposed to find lim y as x approaches infinity.
• September 3rd 2012, 11:32 AM
Preston019
Re: Finding Limits Using L'Hopital's Rule
Quote:

Originally Posted by MaxJasper
$\lim_{x\to \infty } \, \sqrt[x]{\log (x)} = 1$

I need the steps. And that answer is from wolfram alpha. I know because I already checked hoping to find a solution there first.
• September 3rd 2012, 11:59 AM
Plato
Re: Finding Limits Using L'Hopital's Rule
Quote:

Originally Posted by Preston019
I need the steps. And that answer is from wolfram alpha. I know because I already checked hoping to find a solution there first.

Because ${\lim _{x \to \infty }}\frac{{\log (x)}}{x}\mathop = \limits^H {\lim _{x \to \infty }}\frac{{\frac{1}{x}}}{1} = 0$ we know that ${\lim _{x \to \infty }}{\left( {\log (x)} \right)^{\frac{1}{x}}} = {e^0} = 1$
• September 3rd 2012, 12:15 PM
SworD
Re: Finding Limits Using L'Hopital's Rule
Quote:

True. But that's not the answer. We're supposed to find lim y as x approaches infinity. So far, all I have is what lim ln(y) as x approaches infinity equals. I'm supposed to find lim y as x approaches infinity.
But it will give you the answer.If limit ln(y) = 0, then lim y has to equal 1, because you know that as the limit tends to infinity, if ln(y) tends to 0, what does y have to tend to? It will = 1, since ln(1) = 0. Basically raise e to both sides. You are allowed to do that because ln() is continuous at that point.