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Steps to use L'hospital Rule?

  1. Nov 1, 2008 #1
    Hey there..
    I studied limits long time ago ofcourse.. but I used to use an old way in solving them, because the l'hospital rule wasn't allowed:)
    My question is..can someone please help in giving me the steps I should follow in solving a limit using L'hospital's rule?
    Thanks a lot..I really need to know how to solve limits..
  2. jcsd
  3. Nov 1, 2008 #2


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    If you're looking at the limit of f(x)/g(x), and it's in indeterminate form (either f and g go to 0, or f and g go to infinity) you can look at the limit of f'(x)/g'(x) instead, and if that limit exists it equals the limit of f/g. Sometimes you have to do this more than once
  4. Nov 1, 2008 #3


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    If you have a a fraction of the form f(x)/g(x) and f and g separately both go to 0 or both go to infinity (as x goes to a), then
    [tex]\lim_{x\rightarrow a} \frac{f(x)}{g(x)}= \lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}[/tex]

    That's what OfficeShredder said. I want to add that L'Hopital's rule can be used in other cases:
    If we have f(x)g(x) with one of f(x) or g(x) going to 0 and the other to plus or minus infinity, then we can rewrite the problem as either f(x)/(1/g(x)) or g(x)/(1/f(x)) so we have the "0/0" or "[itex]\infty/\infty[/itex]" case.

    If we have F(x)= f(x)g(x) and f(x) and g(x) both go to 0, then we can take the logarithm: ln(F(x))= g(x)ln(f(x)). Now g(x) goes to 0 while ln(f(x)) goes to negative infinity, the previous case. If this new limit is A, then the limit of F is eA.
    If we have f(x)g(x), with f and g both going to
  5. Nov 1, 2008 #4
    Thanx a lot Office_shredder and hallsofIvy..
    u guys were a lot of help..so u mean that if I don't get an indeterminate quantity, I should manipulate the functions to get an indeteminate quantity..aha..
    Thanx a lot again...
  6. Nov 1, 2008 #5


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    No! That means if you don't get an indeterminate quantity, you don't NEED L'Hopitals rule! Just use the quantity you did get.
  7. Nov 2, 2008 #6
    oh ok..
    I get it now..
    Thanx again..
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