Solving L'Hospital's Rule Homework on f(x)+f'(x)=L

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The discussion focuses on solving a homework problem involving L'Hospital's Rule applied to the function f(x) under the conditions that lim_{x->\infty} [f(x)+f'(x)]=L, lim f(x)=L, and lim f'(x)=0. Participants explored various approaches, including substituting f(x) and f'(x) based on the hint provided, which suggests using the expression f(x)=e^{x}f(x)/e^{x}. The consensus is to utilize L'Hospital's Rule on the transformed function to analyze its limits effectively.

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Homework Statement


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


Homework Equations





The Attempt at a Solution


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


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.
 
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Use l'hopital rule on f(x) = [e^xf(x)] / e^x
 
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:
rapple said:

Homework Statement


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


Homework Equations





The Attempt at a Solution


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


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 =]
 
JG89 said:
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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