# Finding the limit of this function

1. Jan 26, 2012

### turbokaz

1. The problem statement, all variables and given/known data

limit as x→3 f(x)= 7/ln(x-2) - 7/(x-3)

2. Relevant equations

3. The attempt at a solution
Cross multiplied to get it into one quotient: 7x-21-7ln(x-2)/ln(x-2)(x-3). Plugged 3 in to get Indeterminate form 0/0. Took derivative of top and bottom then plugged 3 in again and got -6/0. NOT RIGHT. The correct answer is 7/2. I am stuck as to how they got that.

2. Jan 26, 2012

### Dick

You should have gotten 0 for the derivative of the numerator as well. Indicating you would want to take l'Hopital again. Can you show why you didn't?

3. Jan 26, 2012

### turbokaz

okay, I see that I should have gotten 0/0 a second time. But I do L'hopital's again and get (7/(x-2)^2)/ln(x-2)+(x-3/x-2) which after plugging in 3 gives me 7/0.

4. Jan 26, 2012

### Dick

Why do you still have an ln in the denominator after taking the next derivative? You aren't showing your work and you are being sloppy about what you aren't showing.

5. Jan 26, 2012

### turbokaz

From the original equation then. After cross multiplying, I plug in and get 0/0. After round one of L'hopitals, I get (7-(7/(x-2)))/ln(x-2)(x-3). Plug 3 in and get 0/0 again. Derivative of the top gives me (7/(x-2)^2) because 7 goes to 0 and by quotient rule of the second term. I used the product rule on the denominator of ln(x-2)*(x-3) and got ln(x-2)+(1/x-2)*(x-3).

6. Jan 26, 2012

### Dick

Ok, thanks for helping. So why didn't you take the derivative of the denominator in the first round of l'Hopital's? After the FIRST round I've got ln(x-2)+(x-3)/(x-2) in the denominator.

7. Jan 26, 2012

### turbokaz

I don't know why I was being so sloppy. You're right, after round 1 I should have that in the denominator, giving way to 0/0 again. Taking the derivative of top and bottom again leads to (7/(x-2)^2)/ 1/(x-2) + 1/(x-2)^2. Plugged in 3 and got the 7/2. I don't know why my work was so messy but whatever, guess I need to SLOW down. Thanks for your patience.

8. Jan 26, 2012

### Dick

You're welcome. You fixed your own problem. Good work!