# Thread: Limit Problem involving ln

1. ## Limit Problem involving ln

what is lim as x approaches 1 for ln(ln x)/lnx ?

2. Originally Posted by mariam
lim as x approaches 1 for ln(lnx)/lnx
as $\displaystyle x \to 1$ , the numerator approaches $\displaystyle -\infty$ , and the denominator approaches $\displaystyle 0$ ... what does that tell you?

3. i think this means that it does not exist

4. Originally Posted by skeeter
as $\displaystyle x \to 1$ , the numerator approaches $\displaystyle -\infty$ , and the denominator approaches $\displaystyle 0$ ... what does that tell you?
actually, the numerator only approaches $\displaystyle - \infty$ as $\displaystyle x \to 1$ from the right...

Originally Posted by mariam
i think this means that it does not exist
you are correct, but probably not for the reason you think you are. the problem is, as $\displaystyle x \to 1$ from the left, $\displaystyle \ln x$ is negative, and so $\displaystyle \ln ( \ln x )$ would be undefined. As the logarithm function is only defined when its argument is positive.

So since the numerator is undefined as we approach 1 from the left, the entire function is undefined there, and so the limit does not exist for that reason.

Now, $\displaystyle \lim_{x \to 1^+} \frac {\ln ( \ln x)}{\ln x}$ does exist. That should be fun for you to find (but not that difficult--honest)

5. I believe as x approchase 1 from the right, ln(x) equals to negitive infinity.
By using L Hopitals rule and taking the derivative of both numerator and denominator. so as x approchase 1 from the right 1/ln(x) = 1/negative infinity = 0

6. Originally Posted by mariam
I believe as x approchase 1 from the right, ln(x) equals to negitive infinity.
By using L Hopitals rule and taking the derivative of both numerator and denominator. so as x approchase 1 from the right 1/ln(x) = 1/negative infinity = 0
exactly how would you apply L'Hopital's rule here? you need the form to be $\displaystyle \frac 00$ or $\displaystyle \frac {\infty}{\infty}$ to use that. you have neither of those forms here...

7. I am not sure if this make sense. now I know that as x approchase 1 from the right ln(x)= negitive infinity.. so as x approchase to 1 from the right ln(ln(x))/ln(x)= ln(negitive infinity)/negitive infinity.. so now the numerator undifined and the denominator equal to negitive infinity
This all I know

8. Originally Posted by mariam
I am not sure if this make sense. now I know that as x approchase 1 from the right ln(x)= negitive infinity.. so as x approchase to 1 from the right ln(ln(x))/ln(x)= ln(negitive infinity)/negitive infinity.. so now the numerator undifined and the denominator equal to negitive infinity
This all I know
No, $\displaystyle \lim_{x \to 1^+} \ln x = 0$

you have $\displaystyle \lim_{x \to 1^+} \frac {\ln \ln x}{\ln x} \to \frac {\ln 0^+}{0^+} \to \frac {- \infty}0$...loosely speaking.

what can you do with that?

9. Is it this does not exist?

10. Originally Posted by mariam
Is it this does not exist?
No, it exists...

Here's the trick, write it as $\displaystyle \lim_{x \to 1^+} \ln \ln x \cdot \frac 1{\ln x}$

Now what?

11. Trust me I have no clue.

12. Originally Posted by mariam
Trust me I have no clue.
$\displaystyle \lim_{x \to 1^+} \ln \ln x \cdot \frac 1{\ln x} \to - \infty \cdot \infty$