Finding the limit of this function

[turbokaz]

Homework Statement

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

Homework Equations

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.

 

[Dick]

Homework Statement

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

Homework Equations

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.

 

[Dick]

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).

 

[Dick]

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.

 

[Dick]

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.

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