# l'Hopital's Rule

• Oct 21st 2007, 01:03 PM
l'Hopital's Rule
I'm supposed to use l'Hopital's rule to calculate:
lim x-->infinity: [ln(ln(x))]/[xln(x)]

I know I find the derivative of the numerator and the derivative of the denominator.

For the numerator: Dx ln(ln(x)) = 1/(ln(x))
For the denominator (product rule since I have two terms, x, and ln(x) ) Dx xln(x) = (ln(x)) (x(1/x))

I'm not sure if that's correct or not, but I know I have to actually calculate the limit and that doesn't seem possible with the way the equations stands after the first derivative of the numerator and denominator. If somebody could show me what I have to do next that would be great.

Thanks
• Oct 21st 2007, 01:17 PM
ThePerfectHacker
Quote:

I'm supposed to use l'Hopital's rule to calculate:
lim x-->infinity: [ln(ln(x))]/[xln(x)]

I know I find the derivative of the numerator and the derivative of the denominator.

For the numerator: Dx ln(ln(x)) = 1/(ln(x))
For the denominator (product rule since I have two terms, x, and ln(x) ) Dx xln(x) = (ln(x)) (x(1/x))

I'm not sure if that's correct or not, but I know I have to actually calculate the limit and that doesn't seem possible with the way the equations stands after the first derivative of the numerator and denominator. If somebody could show me what I have to do next that would be great.

Thanks

$\displaystyle \lim_{x\to \infty} \frac{\ln (\ln x)}{x\ln x} = \lim_{y\to \infty} \frac{\ln y}{e^y y}=0$.
• Oct 21st 2007, 01:17 PM
galactus
Differentiating the num, one gets $\displaystyle \frac{1}{xln(x)}$

The den, one gets $\displaystyle ln(x)+1$

Then we have $\displaystyle \lim_{x\rightarrow{\infty}}\frac{1}{xln(x)(ln(x)+1 )}$

Now, see the limit?.
• Oct 21st 2007, 01:24 PM
Quote:

Originally Posted by galactus
Differentiating the num, one gets $\displaystyle \frac{1}{xln(x)}$

The den, one gets $\displaystyle ln(x)+1$

Then we have $\displaystyle \lim_{x\rightarrow{\infty}}\frac{1}{xln(x)(ln(x)+1 )}$

Now, see the limit?.

I suppose that the denominator is going to infinty meaning the limit will be 0. However, could you explain how you came to your solutions for the numerator and denominator, I tried to figure them out, but I couldn't see how you worked them out.

Thanks again for all the help.
• Oct 21st 2007, 01:29 PM
galactus
I just differentiated the top and bottom.

Derivative of $\displaystyle ln(ln(x))$ using chain rule.

The derivative of ln(x) is 1/x, so with the chain rule we have $\displaystyle \frac{1}{xln(x)}$

Using product rule, the derivative of xln(x) is $\displaystyle x(\frac{1}{x})+ln(x)=ln(x)+1$

Dividing, we get $\displaystyle \frac{1}{xln(x)(ln(x)+1)}$

So, you are correct, the limit is 0.
• Oct 21st 2007, 01:33 PM
ThePerfectHacker
Here is another non L'Hopital way:
$\displaystyle \left| \frac{\ln (\ln x)}{x\ln x} \right| \leq \frac{\ln x}{x\ln x} = \frac{1}{x} \to 0$
• Oct 21st 2007, 01:38 PM
galactus
L'Hopital is a nice fall back, but I am like PH on this. I like to find them without it if possible. And it's mostly possible. Since you're required to learn L'Hopital, go with it.
• Oct 21st 2007, 01:40 PM
Right, I forgot about the chain rule, thanks a bunch guys.
• Oct 21st 2007, 02:48 PM
While we're on the topic of l'Hopital, I saw a problem in my textbook (with no solution) that looked like this:

lim x--> infinity: (3x-4)/(2x-5)

If I differentiate the numerator and denominator I get 3/2 for the limit as x goes to infinity, would this be correct? It seems kind of simple, but then again, maybe it was an introductory problem to the concept of l'Hopital's rule.

Thanks once again.
• Oct 21st 2007, 02:51 PM
Jhevon
Quote:

While we're on the topic of l'Hopital, I saw a problem in my textbook (with no solution) that looked like this:

lim x--> infinity: (3x-4)/(2x-5)

If I differentiate the numerator and denominator I get 3/2 for the limit as x goes to infinity, would this be correct? It seems kind of simple, but then again, maybe it was an introductory problem to the concept of l'Hopital's rule.

Thanks once again.

yes, the limit is 3/2, and there are several ways to get this without L'Hopital's
• Oct 21st 2007, 02:52 PM
galactus
Yes, that's correct. You certaily don't need the 'Hospital' rule.

Possibly, an easy, early one to introduce...as you said.