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Homework Help: Calculus I - Finding a limit

  1. Sep 14, 2010 #1

    Dembadon

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    Consider the following function:

    [tex]f(x) = 4x^2 - 8x.[/tex]

    Find the limit.

    [tex]
    \lim_{{\Delta}x\rightarrow 0} \frac{f(x+{\Delta}x)-f(x)}{{\Delta}x}
    [/tex]

    Given: The limit exists.


    3. The attempt at a solution

    Since the limit exists, I know that I need to do some algebraic manipulations that will enable me to cancel the [tex]{\Delta}x[/tex] in the denominator.


    Here's what I did first:

    [tex]
    \frac{4(x+{\Delta}x)^2-8(x+{\Delta}x)}{{\Delta}x}
    [/tex]


    After expanding:

    [tex]
    \frac{4(x^2+2x{\Delta}x+{\Delta}x^2)-8(x+{\Delta}x)}{{\Delta}x}
    [/tex]


    After distributing:

    [tex]
    \frac{4x^2+8x{\Delta}x+4{\Delta}x^2-8x-8{\Delta}x}{{\Delta}x}
    [/tex]


    Would my next step be?:

    [tex]
    \frac{(4{\Delta}x^2+8x{\Delta}x-8{\Delta}x)+(4x^2-8x)}{{\Delta}x}
    [/tex]

    ...so that I could pull out the [tex]{\Delta}x[/tex] and cancel it?
     
  2. jcsd
  3. Sep 14, 2010 #2

    Dick

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    Science Advisor
    Homework Helper

    You forgot to subtract the 'f(x)' that's in the numerator of your difference quotient. That's what cancels the stuff that doesn't have a delta-x in it.
     
  4. Sep 14, 2010 #3
    The second group (4x^2-8x) in the numerator does not have a {\Delta}x so you can't completely get rid of the {\Delta}x. Do you have any other attemps?
     
  5. Sep 14, 2010 #4

    Dembadon

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    Gold Member

    What an embarrassingly careless mistake. Thank you, Dick. I've obtained the correct expression for the limit.

    My initial difference quotient should've been:

    [tex]
    \frac{4(x+{\Delta}x)^2-8(x+{\Delta}x)-(4x^2-8x)}{{\Delta}x}
    [/tex]

    Thanks for the help!
     
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