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Really couldn't catch the concept on epsilon and delta in limits

  1. May 26, 2012 #1
    Really couldn't catch the concept on epsilon and delta in limits.

    Let ∂x=x2 - x1
    In finding a gradient the value ∂y is taken at certain value.
    But in finding area using integral, the ∂y is seen to taken as zero. F(x2)=F(x1)

    Maybe one multiplication and the other is division.
    Last edited: May 26, 2012
  2. jcsd
  3. May 26, 2012 #2


    Staff: Mentor

    Re: Limits

    I think you mean Δx = x2 - x1. Δx means "the change in x."
    Δ - uppercase Greek letter delta
    δ - lowercase Greek letter delta

    This is like the difference between D and d in the Roman letters.

    This symbol - ∂ - is used for partial derivatives.

    You probably mean Δy, "the change in y," which would be f(x2) - f(x1).

    The slope of a secant line between two points (x1, f(x1)) and (x2, f(x2)) on a curve is given by

    $$m = \frac{\Delta y}{\Delta x} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$$

    The derivative (also called gradient) is the limit of either fraction above as Δx approaches 0, or as x2 - x1 approaches 0.
    No, not really. The differential in an integral would be dx or dt (or other), not ∂x or ∂t. If you're asking questions about slope and deltas, you're probably getting ahead of yourself in asking about differentials.
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