# Thread: Limit with Indeterminate Form

1. ## Limit with Indeterminate Form

Hi. I have this problem on my homework:

$\displaystyle \lim_{x \to 1} \frac{1}{ln(x)}-\frac{1}{x-1}$

I know it's the indeterminate form $\displaystyle \infty - \infty$. I've tried using l'Hopital's Rule, but it doesn't seem to help. (It just becomes $\displaystyle \lim_{x \to 1} - \frac{1}{xln(x)^2} - \frac{1}{(x-1)^2}$ which isn't any better.)
I've also tried combining the fractions and taking the derivative, which is extraordinarily tedious and also doesn't help. I know by graphing that the limit is supposed to be 1/2, but I can't figure out a way to make that work. Is there something I'm missing? I haven't been this bothered by a math problem in a long time. Any help you can offer would be greatly appreciated. Thanks!

2. Hello,
Originally Posted by capandbells
Hi. I have this problem on my homework:

$\displaystyle \lim_{x \to 1} \frac{1}{ln(x)}-\frac{1}{x-1}$

I know it's the indeterminate form $\displaystyle \infty - \infty$. I've tried using l'Hopital's Rule, but it doesn't seem to help. (It just becomes $\displaystyle \lim_{x \to 1} - \frac{1}{xln(x)^2} - \frac{1}{(x-1)^2}$ which isn't any better.)
I've also tried combining the fractions and taking the derivative, which is extraordinarily tedious and also doesn't help. I know by graphing that the limit is supposed to be 1/2, but I can't figure out a way to make that work. Is there something I'm missing? I haven't been this bothered by a math problem in a long time. Any help you can offer would be greatly appreciated. Thanks!
You cannot apply l'Hospital's rule in this form

Let t=x-1

The limit is now : $\displaystyle \lim_{t \to 0} \frac{1}{\ln(1+t)}-\frac 1t$

$\displaystyle =\lim_{t \to 0} \frac{t-\ln(1+t)}{t \ln(1+t)}$

now apply the rule twice and you'll get the result (1/2)

3. Wow, thank you for your quick response! I feel dumb, but I appreciate it.