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L'Hospitals Rule

  • Thread starter rapple
  • Start date
  • #1
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Homework Statement


Given f is differentiable on (0,[tex]\infty[/tex])
Given [tex]lim_{x->[tex]\infty[/tex]}[/tex] [f(x)+f'(x)]=L
S.T lim f(x)=L and lim f'(x)=0
Hint f(x)=e[tex]^{x}[/tex]f(x)/e[tex]^{x}[/tex]


Homework Equations





The Attempt at a Solution


A Lim [tex]_{x->[tex]\infty[/tex]}[/tex] [f(x)+f'(x)]=L
Then for some [tex]\epsilon[/tex]>0
|f(x)+f'(x)-L|<[tex]\epsilon[/tex]


Tried different approaches by substituting for f(x) and f'(x) based on the hint. But did not help. I tried to get it to a L/infinity form so f'(x)=0 but could not.
 

Answers and Replies

  • #2
726
1
Use l'hopital rule on f(x) = [e^xf(x)] / e^x
 
  • #3
164
2
its unclear what you're actually asking. f(x)=exp(x)*f(x)*exp(-x) makes no sense to me.

i'd try proof by contradiction and show all such conditions cannot hold.

lets say f(x) is merely an increasing function,
(1) is it possible that a bounded f(x) has nonzero slope everywhere?
(2) is it possible that an unbounded f(x) eventually has zero slope?

[edit: oh okay yeah, use the hint.]
 
Last edited:
  • #4
djeitnstine
Gold Member
614
0

Homework Statement


Given f is differentiable on (0,[tex]\infty[/tex])
Given [tex]lim_{x->[tex]\infty[/tex]}[/tex] [f(x)+f'(x)]=L
S.T lim f(x)=L and lim f'(x)=0
Hint f(x)=e[tex]^{x}[/tex]f(x)/e[tex]^{x}[/tex]


Homework Equations





The Attempt at a Solution


A Lim [tex]_{x->[tex]\infty[/tex]}[/tex] [f(x)+f'(x)]=L
Then for some [tex]\epsilon[/tex]>0
|f(x)+f'(x)-L|<[tex]\epsilon[/tex]


Tried different approaches by substituting for f(x) and f'(x) based on the hint. But did not help. I tried to get it to a L/infinity form so f'(x)=0 but could not.
Just a side note, you do not have to use separate wraps for Latex, simply use one so that we can understand it better =]
 
  • #5
25
0
Use l'hopital rule on f(x) = [e^xf(x)] / e^x
How?

I can see that x->infinity, e^xf(x)/e^x is of the infinity . limx->inf f(x)/infinity. Since we don't know anything about f(x) except it s continuous and differentiable on (0,infnty), can i make the conclusion that it is not= 0 hence is of the infnty/infnty form.
Proceeding with that thought:
lim x->infnty f'(x)= limx->infnty [(e^xf(x) +e^x)f'(x))/e^x]. Now this is infnty . L/infnty form. What do I do after.I can see that lim x->infnty f'(x) Not=L but how do I show it is 0.
 

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