Limit and l'hopital's rule

  • #1
lim as x approaches 4 from the right of [3(x-4)]^(x-4)

when you plug in 4 it becomes a 0^0. So I attempted to manipulate it so it would become a 0/0 form so I could l'hopital it. (sorry if I keep spelling it wrong)

so lim(x-4)ln(3x-12)=lny
then I l'hopitaled it:
lim (x-4)(1/3x-12)*3+ln(3x-12)=lny
lim 1+ln(3x-12)=lny
lim e + (3x-12)=y
so y=e

I do not know if my logic makes sense or not. If anyone sees any flaws in my logic, I would really like to know.
 

Answers and Replies

  • #2
201
0
lim as x approaches 4 from the right of [3(x-4)]^(x-4)

when you plug in 4 it becomes a 0^0. So I attempted to manipulate it so it would become a 0/0 form so I could l'hopital it. (sorry if I keep spelling it wrong)

so lim(x-4)ln(3x-12)=lny
then I l'hopitaled it:
lim (x-4)(1/3x-12)*3+ln(3x-12)=lny
lim 1+ln(3x-12)=lny
lim e + (3x-12)=y
so y=e

I do not know if my logic makes sense or not. If anyone sees any flaws in my logic, I would really like to know.

why don't you try this:

lim[tex]\frac{ln(3x-12)}{1/(x-4)}[/tex]

lhopital this until you reach something that is defined.
 
  • #3
hmmm ok so I get that the limit instead aproaches 0, however I look on my calculator and it should instead approach -infinity.
I had that: ln(3x-12)/1/x-4
so l'hopital it to get:
1/3x-12 * 3/(1/(x-4)^2)=0/0
so I l'hopitaled it again:
-1/(x-4)^2/(-2(x-4)/(x-4)^4)
so that the limit approaches 0. So where did I go wrong?
 
  • #4
Dick
Science Advisor
Homework Helper
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Use some more parentheses when you write this stuff, ok? It's hard to figure out whether 1/3x-12 is supposed to (1/(3x))-12 or (1/3)x-12 or what you really want which is 1/(3x-12). Your first l'Hopital form isn't 0/0. It's infinity/infinity. But if you do some algebra on it and simplify, you'll see you don't have l'Hopital it again. And if you are getting 0 that means ln of your original limit is 0. What's the original limit?
 
  • #5
ok so if its: 1/(3x-12)*3/[1/(x-4)^2]
I simplified it to get: 1/(x-4)/(1/(x-4)^2)
so x-4 basically
which gets me...infinity?
 
  • #6
Dick
Science Advisor
Homework Helper
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ok so if its: 1/(3x-12)*3/[1/(x-4)^2]
I simplified it to get: 1/(x-4)/(1/(x-4)^2)
so x-4 basically
which gets me...infinity?

x approaches 4. Is that really infinity?
 
  • #7
0...gah I don't get this. Since on my calc it looks like it should approach 1
 
  • #8
201
0
0...gah I don't get this. Since on my calc it looks like it should approach 1

its
lny=0
when x --> 4+

however y approaches 1

lny=0
e^(lny)=e^0
y=1
 
  • #9
HallsofIvy
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lim as x approaches 4 from the right of [3(x-4)]^(x-4)

when you plug in 4 it becomes a 0^0. So I attempted to manipulate it so it would become a 0/0 form so I could l'hopital it. (sorry if I keep spelling it wrong)

so lim(x-4)ln(3x-12)=lny
then I l'hopitaled it:
lim (x-4)(1/3x-12)*3+ln(3x-12)=lny
This is NOT L'Hopital. On the left you have just the limit of the derivative of ln y.

lim 1+ln(3x-12)=lny
lim e + (3x-12)=y
so y=e

I do not know if my logic makes sense or not. If anyone sees any flaws in my logic, I would really like to know.
 

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