Finding the limit of this function

turbokaz

Homework Statement

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

Homework Helper

Homework Statement

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

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?

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.

Homework Helper
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.

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.

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).

Homework Helper
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).

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.

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.