# Limit (L'hopitals)

1. Mar 2, 2013

### whatlifeforme

1. The problem statement, all variables and given/known data
evaluate the limit using l'hopitals rule.

2. Relevant equations
limit x->9+ (1/(x-9) - 1/ln(x-8))

3. The attempt at a solution
I tried taking the derivative of the numerator and denominator in each, getting 0 as an answer, but it wasn't correct.

2. Mar 2, 2013

### CAF123

Reexpress that as a single fraction then |'Hopital's Rule becomes valid.

3. Mar 3, 2013

### whatlifeforme

ln(x-8) - x - 9
------------------ 0/9 ---> should be zero right?
x - 9ln(x-8)

anyways, apply l'hopitals (take derivative of numerator and denominator):

(1/x-8) - 1
------------ = 0/-8 = 0
1 - 9(1/x-8)

answer should be, but i can't figure out how: (-1/2)

4. Mar 3, 2013

### eumyang

Be careful with your algebra. I'm getting
$$\frac{\ln (x-8) - x + 9}{(x - 9)\ln(x - 8)}$$
Of course, as x approaches 9 from the right the fraction goes to 0/0.

5. Mar 3, 2013

### Dick

I really wish you would write things using more parentheses where needed. You have (log(x-8)-(x-9))/((x-9)*log(x-8)). The denominator is a product. Use the product rule. You'll should get another 0/0 limit. That just means you should use l'Hopital again.

6. Mar 3, 2013

### whatlifeforme

now i'm getting 1 as the answer, after correcting my algebra, and applying l'hopitals again.

(1/(-x+8)^2)
lim(x->9+ ------------------
(x-9)(1/(-x+8)^2) + (1/(x-8)) - 9ln(x-8)

= (1/1)/1

7. Mar 3, 2013

### eumyang

You should also show all steps in your work, not just the last one. It's harder to detect where you went wrong. You should also learn LaTeX -- it's very hard to read what you wrote above.

What did you get when you took the first derivative? Before taking the second derivative, you should simplify the expression by multiplying top and bottom by (x - 8). That way you won't have to deal with complex fractions.

8. Mar 3, 2013

### whatlifeforme

sorry i cna't figure the latex out.

1. limit x->9+ (1/(x-9) - 1/ln(x-8))

2. ln(x-8) - x + 9
----------------- Common denominator. (0/0)
(x-9) * (ln(x-8))

3. (1/(x-8)) - 1
----------------- derivatives of numerator and denominator. using product rule in denominator. (0/0)
(x-9) * (1/(x-8)) + 1(1/ln(x-8))

4. (-1/(x-8)^2)
--------------------- derivatives of numerator and denominator. using product rule in denominator. (1/1)
(x-9)* (-1/(x-8)^2) + (1/(x-8) + (-1/(x-8)*ln(x-8)^2)

Last edited: Mar 3, 2013
9. Mar 3, 2013

### eumyang

If I can do it, so can you!

This is wrong.

After you fix step #3, as I said previously, before you take the derivative again, multiply the numerator and denominator by (x-8), so that you won't have to deal with complex fractions.

10. Mar 3, 2013

### whatlifeforme

what's wrong with it?

11. Mar 3, 2013

### Dick

eumyang marked the problem in red. How did you 1/ln(x-8) in the denominator. That's wrong.

12. Mar 4, 2013

### whatlifeforme

i'm sorry; it should be.

(x-9) * (1/(x-8)) + 1(ln(x-8))

13. Mar 4, 2013

### Dick

Ok, so you've fixed that. So it's 0/0. Now take another derivative.