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Limit using L'hopital's rule

  1. Jul 1, 2008 #1
    1. The problem statement, all variables and given/known data

    find the limit as x approaches 1- of [(1-x^2)^1/2] / [(1-x^3)^1/2] aka root(1-x^2)/root(1-x^3)

    2. Relevant equations

    3. The attempt at a solution

    Well I don't really get how to solve this limit using l'hopital's rule if i differentiate both top and bottom i get [1 / 2root(1-x^2)] / [1 / 2root(1-x^3) which is now in the form inf / inf , or if i cross multipl it assumes the same form of my original limit statement with numrator and denominator switched, and if i differentiate top ad bottom again, it goes back to its original form...So basically it seems like one endless loop.
    The only wayI can think to get the answer is to remove the square roots (since they apply to both the top and bottom of the fraction), find the limit as x approaches 1 of (1-x^2) / (1-x^3) which is just 2/3. So the original limit must have been root (2/3)
    Now that seems to match the answer in the back of the book, but my first instinct was to apply L'hopital's rule even with the root signs, and that didn't work out at all. Can someone explain to me if there is a way to solve this problem by applying L'hopital's rule even with the roots, and, if not, is there other similar situations n which i have to be sneakier in applying L'hopital's rule?
  2. jcsd
  3. Jul 1, 2008 #2


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    Homework Helper

    There's part of your difficulty right there -- you're forgetting about the Chain Rule. There should be another factor in both the numerator and the denominator...

    Try factoring inside the radical of the original expression first: the numerator is a difference of two squares and the denominator is a difference of two cubes. You should find that you can eliminate the obnoxious (1 - x) factor. (Of course, now it's not an indeterminate ratio. Are you required to use L'Hopital on this?)
    Last edited: Jul 1, 2008
  4. Jul 1, 2008 #3
    In the first part, you're forgetting to multiply by the derivative of the function inside the radicals. You were right in trying to pass the limit operator over the square root function, that is a not so well known theorem. If you have a limit operator outside of a function, which does no involve the variable, the limit operator may be safely moved across the function. It's what's used in proving the famous e limit,
    [tex]e=\lim_{x\rightarrow \infty}\left(1+\frac{1}{x}\right)^x[/tex].
    And to show you an example,
    [tex]\lim_{x\rightarrow c}\left(\sin f(x)\right)=\sin\left(\lim_{x\rightarrow c}f(x)\right)[/tex]
    I can't think of any other tricky situations where you'd have to come up with something creative with L'hopital's rule. Just know that if you get an indeterminate form after one iteration, try another. Sometimes 3 or 4 are needed.
    Hope that helps.
  5. Jul 1, 2008 #4
    ohhhh crap haha I can't believe I forgot thatpart :) thank u :)
  6. Jul 1, 2008 #5


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    I'm not clear on what you mean by "does not involve the variable" since both of those examples certainly do. In any case, that is simply the definition of "continuous function". As long as g(x) is continuous at f(a) then
    [tex]\lim_{x\rightarrow a} g(f(x))= g(\lim_{x\rightarrow a} f(a))[/tex]

  7. Jul 1, 2008 #6
    If the function you were trying to pass the limit operator over involved the variable it wouldn't work, for instance,
    [tex]\lim_{x\rightarrow c} x \sin\left( f(x)\right)\neq x\sin\left(\lim_{x\rightarrow c} f(x)\right)[/tex].
    in your example, g(f(x))=x Sin[f(x)]. So g(x) cannot involve the variable itself. So basically you can only pass the limit operator over other operators i.e sin, cos, exp, log.
  8. Jul 1, 2008 #7


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    That "interpretation" occured to me a little later! In that case, again if g(x,y) is continuous at (c, f(c)),
    [tex]\lim_{x\rightarrow c}g(x,f(x))= g(c, \lim_{x\rightarrow c}f(x))[/tex]
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