# l'Hopital's Rule

• October 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
• October 21st 2007, 01:17 PM
ThePerfectHacker
Quote:

Originally Posted by Adrian
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

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

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

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

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

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

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

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

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.
• October 21st 2007, 01:29 PM
galactus
I just differentiated the top and bottom.

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

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

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

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

So, you are correct, the limit is 0.
• October 21st 2007, 01:33 PM
ThePerfectHacker
Here is another non L'Hopital way:
$\left| \frac{\ln (\ln x)}{x\ln x} \right| \leq \frac{\ln x}{x\ln x} = \frac{1}{x} \to 0$
• October 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.
• October 21st 2007, 01:40 PM
Right, I forgot about the chain rule, thanks a bunch guys.
• October 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.
• October 21st 2007, 02:51 PM
Jhevon
Quote:

Originally Posted by Adrian
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
• October 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.