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Limit problem (by definition I think)

  1. Apr 12, 2016 #1
    1. The problem statement, all variables and given/known data
    ##f(x)=12x^2-5##
    The value of ##\lim_{h\rightarrow 0}\frac{f(x+2h)-f(x-3h)}{6h}## is ...

    A. 8x
    B. 10x
    C. 12x
    D.18x
    E. 24x

    2. Relevant equations

    ##f'(x) = \lim_{h->0}\frac{f(x+h)-f(x))}{h}##

    3. The attempt at a solution

    Looking at the problem question, it seems that it's kinda like the definition of derivative
    But, it's different.
    This is my attempt.

    ##\lim_{h\rightarrow 0}\frac{f(x+2h)-f(x)}{2h}*\frac{2h}{6h}-\frac{f(x-3h)-f(x)}{3h}*\frac{3h}{6h}##
    ##= f'(x) * 1/3 - f'(x) * 1/2##
    ##= f'(x)*(-1/6)##
    ##= 24x (-1/6)##
    ##= -4x##

    But, it's not in the options. Please help
     
  2. jcsd
  3. Apr 12, 2016 #2

    PeroK

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    The derivative is not relevant. Why not just evaluate the limit?

    I didn't check where you went wrong, but here is no need to complicate things. Just evaluate the limit as you see it.

    PS The answer I get isn't on the list.
     
    Last edited: Apr 12, 2016
  4. Apr 12, 2016 #3

    Ssnow

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    In order to have the derivative you must have ##+\frac{f(x-3h)-f(x)}{-3h}## so you have ##24x(1/3+1/2)## ...
     
  5. Apr 12, 2016 #4
    Hmm.. By just evaluating the limit, I get the answer 20x as the answer

    Yeah, I missed that one
    ##24x (\frac{1}{3}+\frac{1}{2}) = 24 * \frac{5}{6} = 20x##

    Thanks for your help @PeroK and @Ssnow !
     
  6. Apr 12, 2016 #5

    SammyS

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    Are you sure there's not a typo? It could be either of the following.
    ##\displaystyle \frac{f(x+2h)-f(x-3h)}{5h}##​
    .
    ##\displaystyle \frac{f(x+3h)-f(x-3h)}{6h}##​

    Both give answer E.
     
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