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Can someone verify my answer for the problem in my post below

  1. Aug 28, 2005 #1
    ok, so I worked out the problem and got 2x+a... is this correct? For some reason I dont think it is, but I thought I worked it out right... I just want to know if I'm on the right track so I dont do all 50 problems wrong
  2. jcsd
  3. Aug 28, 2005 #2
    which problem?
  4. Aug 29, 2005 #3
    sorry, i guess I posted a different problem below...

    lim ( (x+∆x)^2 - x^2 ) / x
    ∆x approaches 0

    Sorry about that
  5. Aug 29, 2005 #4


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    Do you mean
    lim ( (x+∆x)^2 - x^2 ) / ∆x
    ∆x approaches 0?

    If so, you get 2x. Otherwise, if you meant the problem as you typed it, you get 0 (or undefined if x=0).
  6. Aug 30, 2005 #5
    Your problem is a verification of the definition of derivative.
    In particular [tex]\frac{d}{dx}\left[x^2 \right] = \lim _{\Delta x\rightarrow 0} \frac{(x+\Delta x)^2 - x^2}{\Delta x} =
    \lim _{\Delta x\rightarrow 0} \frac{x^2 + 2x\Delta x + (\Delta x)^2 - x^2}{\Delta x} = \lim _{\Delta x\rightarrow 0} \frac{2x\Delta x + (\Delta x)^2 }{\Delta x} = \lim _{\Delta x\rightarrow 0} (2x+ \Delta x) = 2x [/tex]

    Notice that in general (for other functions) both right and left limit have to exist in order to say that the derivative exists. In this case you have a simple polinomyal problem and this is not an issue.

    If you do not like the notation [tex] \Delta x[/tex] for a small number, you can use the equivalent notation:

    [tex]\frac{d}{dx}\left[x^2 \right] = \lim _{\varepsilon \rightarrow 0} \frac{(x+\varepsilon)^2 - x^2}{\varepsilon} =
    \lim _{\varepsilon\rightarrow 0} \frac{x^2 + 2x\varepsilon + (\varepsilon)^2 - x^2}{\varepsilon} = \lim _{\varepsilon\rightarrow 0} \frac{2x\varepsilon + (\varepsilon)^2 }{\varepsilon} = \lim _{\varepsilon\rightarrow 0} (2x+ \varepsilon) = 2x [/tex]
    Last edited: Aug 30, 2005
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