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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


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    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.
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