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Small Limits Question

  1. Oct 7, 2003 #1
    [SOLVED] Small Limits Question

    Hi There,
    I have this Simple Question In Limits :

    If limx->3 (f(x) - 2) / (x - 3) = 7
    Then limx->3 (x2f(x) - 18) / (x-3) = ??

    I solved the question this way :
    Since the denominator equals Zero , and limit exists , then the numerator equals zero .
    [4]f(x) - 2 = 0 ---> f(x) = 2

    limx->3 (x2f(x) - 18) / (x-3) =
    limx->3 (2x2 - 18) / (x-3) =
    limx->3 2(x2- 9) / (x-3) =
    limx->3 2(x-3) (x+3) / (x-3) =
    limx->3 2(x+3) =
    2 (3+3) = 12

    But I also solved in this way :

    limx->3 (x2f(x) - 18) / (x-3) =
    limx->3 (x2f(x) - 2x2 + 2x2 - 18) / (x-3)
    limx->3 (x2f(x) + (2x2) /(x-3) - 2x2 - 18) / (x-3)
    limx->3 (x2) (f(x) + 2) /(x-3) - 2(x2 - 9) / (x-3)
    limx->3 (x2) (f(x) + 2) /(x-3) - 2(x2 - 9) / (x-3)
    limx->3 (x2) 7 - limx->3 2(x-3)(x+3) / (x-3)
    limx->3 (x2) 7 - limx->3 2(x-3)(x+3) / (x-3)
    32 * 7 + limx->3 (x+3)
    (9 * 7) + (3 + 3) = 63 + 12 = 75


    What's the wrong with the first One ?
    Please help.
     
  2. jcsd
  3. Oct 7, 2003 #2
    I'm not sure about this reasoning. Your limit looks very similar to a difference quotient for derivatives. You could write limx-->a(f(x)-f(a))/(x-a)=f'(a). In this case, f'(3)=7, f(3)=2. You want to find limx-->ax2*f'(a)=a2f'(a)=9*7=63
     
  4. Oct 7, 2003 #3
    Hello, Zargawee!

    I found an answer. (Please check my work!)

    We are given: limx->3 [f(x) - 2]/[x - 3] = 7

    We're asked to find: limx->3[x2f(x) - 18]/[x - 3]

    In the numerator, add 2x2 and subtract 2x2:
    x2f(x) - 18 + 2x2 - 2x2 = x2[f(x) - 2] + 2(x2 - 9)

    We have: x2[f(x) - 2]/(x - 3) + 2(x2 - 9)/(x - 3)

    The second term reduces: 2(x - 3)(x + 3)/(x - 3) = 2(x + 3)

    Then we have: x2[f(x) - 2)/(x - 3) + 2(x + 3)

    Taking limits, we have:
    limx->3(x^2) * limx->3 [f(x) - 2]/(x - 3) + limx->3 2(x + 3)

    We are given that the middle limit is 7.

    Therefore, the answer is: (32)(7) + 2(6) = 75
     
    Last edited by a moderator: Oct 7, 2003
  5. Oct 7, 2003 #4

    Hurkyl

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    Gold Member

    That's not quite right; what is true is that

    limx→3 f(x) = 2


    Soroban's answer looks right.
     
  6. Oct 7, 2003 #5
    <--- ashamed
     
  7. Oct 10, 2003 #6
    Thanks all,
    But I think that I solved the question as the same as Soroban solved it (Look at my first post)

    Thanks again.
     
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