- #1
Ad123q
- 19
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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?
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?