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Finding the limit 0/0

  1. Jun 17, 2015 #1
    I am really struggling with limits at the moment. Any help would be great! Thanks to anyone in advance if they take the time to read the rest of this.
    Basically i am struggling with finding the limit when using direct substitution provides 0/0
    I (think) am fine with limits that involve quadratic equations, you factorise which should allow some canceling out, and direct substitution should provide an answer.

    Just to check
    Lim t->(-1) [(1/(t+1)) - (1/(t^2+3t+2)]
    factorizing the t^2+3t+2 i can cancel out to 1/(t+2) which with direct substitution the limit = 1/1 =1

    but my problem is when i encounter limits such as
    lim h->0 (sqrt(2+2h)-sqrt(2)) / h

    I did try multiplying it by the conjugate (sqrt(2+2h)-sqrt(2)) / (sqrt(2+2h)+sqrt(2)) but this gave me 2/sqrt(2) and putting the original into worlfram alpha the answer is apparently 1/sqrt(2)

    I'm just not sure on the procedure when dealing with this king of limit. any help would be greatly appreciated!
    Thanks
     
  2. jcsd
  3. Jun 17, 2015 #2

    micromass

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    Tell us how you obtained ##\frac{2}{\sqrt{2}}##. Your method is good, but there is likely a computation error.
     
  4. Jun 17, 2015 #3
    Hopefully this will be okay to follow apologies if not, not sure how to write equations on here yet.
    >>
    [sqrt(2+2h) - sqrt(2)]/h . [sqrt(2+2h)+sqrt(2)]/sqrt(2+2h)+sqrt(2)
    >>
    2+2h+[sqrt(2)sqrt(2+2h)]-[sqrt(2)sqrt(2+2h)]-2
    h.sqrt(2+2h)+sqrt(2)
    >>
    cancels to
    2h
    h.[sqrt(2+2h)+sqrt(2)]
    >>
    giving, as h->0 & canceling the h's
    2
    sqrt(2+2.0)+sqrt(2)
    Ive realised my error, thanks a lot i will continue just in case
    >>
    2/sqrt(2)+sqrt(2) = sqrt(2)/2

    just for clarification, is it a rule of thumb in this sort of limit to multiply by the conjugate?
     
  5. Jun 17, 2015 #4

    micromass

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    Nice work!

    Yes, it is a trick that works in a lot of cases.
     
  6. Jun 17, 2015 #5
    Thanks! Much appreciated.
     
  7. Jun 17, 2015 #6

    WWGD

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    Have you tried setting ##2h=u## to make the expression look more like a derivative?
     
  8. Jun 18, 2015 #7

    SammyS

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    It looks like it could be a derivative just as it is.

    ƒ'(1) , where ƒ(x) = √(2x) .
     
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