L'Hopitals Rule Conditions

1. Feb 28, 2009

Ad123q

Hi,

I'm slightly confused about one aspect of the conditions for applying L'Hopital's rule.

N.b. apologies in advance for the lack of LaTeX.

L'Hopitals rule:
Let f,g:(a,b) → R be differentiable and let c ε (a,b) be such that f(c)=g(c)=0 and g'(x)≠0 for x≠c.
Then lim(x→ c)[f(x)/g(x)] = lim(x→ c)[f'(x)/g'(x)], provided latter limit exists.

I have lim(x→ 2)[(x²+x-6)/(x²-x-2)] = lim(x→ 2)[(2x+1)/(2x-1)]

I compute this to be 5/3 (which is correct) by L'Hopital's rule.
My point is that this function satisfies f(2)=g(2)=0, but does NOT satisfy g'(x)≠0 for x≠2, as the solution of g'(x)=2x-1=0 is x=0.5; i.e. g'(x)=0 for x=0.5, and 0.5≠c(=2).

So why are we able to apply L'Hopital in this case?

2. Feb 28, 2009

yyat

If take (a,b)=(1,3), then the requirements are satisfied. The point is that you need g'(x) nonzero only "close" to c, i.e. on some open interval containing c.

3. Mar 1, 2009

Overlord

I think you may have stated the conditions incorrectly. In particular, "g'(x)≠0 for x≠c", I believe, should be "g'(x)≠0 for x=c", since that is all that is necessary for the limit of f'(x)/g'(x) to exist (which seems to be consistent with the definitions I've checked from other sources, then if it still doesn't exist, you can apply it again, where it may eventually work).

I might be wrong though. Where did you get the definition of L'Hospital's Rule from?

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