1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lim of this function using L'Hopitals rule multiple times

  1. Sep 15, 2011 #1
    1. The problem statement, all variables and given/known data

    lim as x approached 2 from above of:

    (x1/2 - 21/2 + (x - 2)1/2)/(x2 - 4)1/2

    2. Relevant equations

    L'Hopitals rule...

    3. The attempt at a solution

    I tried susbtituting in 2 and found that I got 0/0 thus it was possible to use L'hopitals. I then got inf/inf, so I used it again, but now I keep getting inf/inf. So what do I do? I'm stumped. I tried factorising but the + and - signs prevented me.
     
    Last edited: Sep 15, 2011
  2. jcsd
  3. Sep 15, 2011 #2
    The reason you keep getting indeterminate form is probably because the pesky power of 1/2. Assuming the limit exists, instead of looking for the limit of f(x) where f(x) is your function there, consider instead looking for the limit of the *square* of f(x).
     
  4. Sep 15, 2011 #3
    But wont that give us a different limit? since the function squared is essentially a different function to the original.
     
  5. Sep 15, 2011 #4
    Of course you would then need to take the square root at the end...
     
  6. Sep 15, 2011 #5
    To be more explicit, if the limit of f and g exist, then lim(f g)=(lim f)(lim g). In particular, lim(f^2)=(limf)^2.
     
  7. Sep 15, 2011 #6

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It follows from the fact that for any two functions [itex]f(x)[/itex] and [itex]g(x)[/itex] such that the limits

    [tex]\lim_{x\to x_0}f(x)\text{ and } \lim_{x\to x_0}g(x)[/tex]

    exist, then

    [tex]\lim_{x\to x_0}[f(x) g(x)] = [\lim_{x\to x_0}f(x)][\lim_{x\to x_0}g(x)][/tex]

    Edit: Seems that I was beaten to it!
     
  8. Sep 15, 2011 #7
    I tried this and the resulting functions again gave me 0/0 so I again used L'hopitals and after length algebra this is the expression I got:

    {2 - (2/x)0.5 + [2(2(x3 - 3x2 + 3x - 2))0.5 - (x2 - 2x)0.5]/ [2x3 - 8x2 + 8x]}

    What do I do now, substituting in 2 gives <1 + inf> where inf comes from the third expression in the equation which gives "(0 - 0)/0". SO what now?
     
  9. Sep 15, 2011 #8
    Now check again. The square of your function should give something nice. You only need to apply L'Hopital *once* to get the answer. In particular your denominator x^2-4 now just becomes 2x. So when x goes to 2, it does not give trouble.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Lim of this function using L'Hopitals rule multiple times
Loading...