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Limit of Delta x question

  1. Jan 18, 2010 #1
    1. The problem statement, all variables and given/known data

    Let [tex]V=x^3[/tex]

    Find [tex]dV[/tex] and [tex]\Delta V[/tex].

    Show that for very small values of [tex]x[/tex] , the difference

    [tex]\Delta V - dV[/tex]

    is very small in the sense that there exists [tex]\varepsilon[/tex] such that

    [tex]\Delta V - dV = \varepsilon \Delta x[/tex],

    where [tex]\varepsilon \to 0[/tex] as [tex]\Delta x \to 0[/tex].

    2. Relevant equations

    [tex]dV = 3x^2 dx[/tex]

    [tex]\Delta V = 3x^2 \Delta x + 3x (\Delta x)^2 + (\Delta x)^3[/tex]

    3. The attempt at a solution

    I worked it down to

    [tex]\varepsilon = 3x \Delta x + (\Delta x)^2 + 3x^2 \left(1 - \frac{dx}{\Delta x} \right)[/tex]

    Can I say that [tex]\lim_{\Delta x \to 0} \Delta x = dx[/tex] so that

    [tex]\lim_{\Delta x \to 0} \varepsilon = 3x(0) + (0)^2 + 3x^2(1-1) = 0[/tex] ?
  2. jcsd
  3. Jan 18, 2010 #2


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    You basically have it. You need to note that [itex]\Delta x[/itex] and [itex]dx[/itex] are the same thing, regardless of limits. Using your equations:

    dV = 3x^2 dx [= 3x^2\Delta x]

    \Delta V = 3x^2 \Delta x + 3x (\Delta x)^2 + (\Delta x)^3

    you have:

    \Delta V-dV= 3x (\Delta x)^2 + (\Delta x)^3 = (\Delta x)(3x\Delta x + (\Delta x)^2))= \Delta x (\varepsilon)

    Does this [itex]\varepsilon[/itex] work? Note that you don't need to take any limits to answer the question. You just need to observe that [itex]\varepsilon \rightarrow 0[/itex] as [itex]\Delta x[/itex] does.
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