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Difference between a limit and a derivative?

  1. Nov 4, 2013 #1
    is it that limit can be taken for something approaching to any value while derivative is limit for the value of that thing approaching to zero?
     
  2. jcsd
  3. Nov 4, 2013 #2
    The derivative is a specific limit, namely:

    lim(h->0) (f(x+h) - f(x))/h

    This can also be expressed as:

    lim(x->a) (f(x) - f(a))/(x-a)

    Any limit that does not always give you the same result as this limit is not a derivative.

    Conceptually, the derivative is the slope of the tangent line, and is exactly the same form as the slope formula for a straight line:

    (y2 - y1)/(x2-x1)

    or

    (y(x2) - y(x1))/(x2-x1)

    or

    Δy/Δx

    or particularly

    dy/dx
     
  4. Nov 4, 2013 #3

    jedishrfu

    Staff: Mentor

    The wikipedia article talks about how limits are used to rigorously define a derivative:

    http://en.wikipedia.org/wiki/Derivative

    When Calculus was first invented, many mathematicians were skeptical of its results until the concept of a limit was applied to more clearly demonstrate the correctness of the results.
     
  5. Nov 4, 2013 #4
    Above all else, as my differential equations professor put it, "a derivative is a rate of change is a rate of change is a rate of change". Be intimately familiar with this fact, and the derivative will always be your friend.
     
  6. Nov 4, 2013 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No. They are much more different than that. The limit of a function, f, as "x approaches a" tells us what value (if it exists) the function takes on arbitrarly close to x= a but NOT equal to it. It is allowed to have a= 0. That does NOT have a separate name! For example, the limit as x goes to 0, of f(x)= 3x+ 2 is just 3(0)+ 2= 2 as we could see by drawing a graph and seeing that, the close x gets to 0, the closer f(x) gets to f(0)= 2. But if we were to define g(x)= 2x- 1 if x> 0, g(0)= 2, g(x)= x^2+ x- 1 if x< 0, and graph y= g(x), we would see that, although g(0)= 2, values of x close to 0 but not equal to it have values of g(x) close to -1, not 2: [itex]\lim_{x\to 0} g(x)= -1[/itex].

    The derivative of a function, f, at x= a, is completely different. It is the "slope of the tangent line" to the graph of y= f(x) at x= a. For example, if f(x)= 3x+ 2, its graph is a straight line. Its "tangent" is just itself so its derivative, at any a, is its slope 3. If we were to draw the graph of y= g(x)= x^2+ x- 1, we would see that it is NOT a straight line so has differernt tangents, with different slopes, at different values of x. At x= 0, the tangent line y= x- 1 which has slope 1 so the derivative of g at x= 0 is 1. But at x= 1, its tangent line is given by y= 3x+ 1 which has slope 3 so the derivative of g at x= 1 is 3.

    What is confusing is that we use a limit to find the derivative. If we are given the function g(x)= x^2+ x- 1, we can see that its value at x= 1 is g(1)= 1+ 1- 1= 1. If we were to look at a second point, just slightly different, say x= 1+ h, we would find that g(1+h)= (1+h)^2+ (1+h)- 1= 1+ 2h+ h^2+ 1+ h- 1= 1+ 3h+ h^2. That is. we have two nearby points on the graph of y= g(x), (1, 1) and (1+h, 1+ 3h+h^2). We can find the slope of the line between these two points (called a "secant line" like a secant in a circle) using the "difference quotient. The difference in y values is (1+ 3h+ h^2)- (1)= h^2+ 3h while the difference in x values is 1+h- 1= h. The quotient of those two differences, the slope of the line through (1, 1) and (1+h, + 3h+ h^2) is (h^2+ 3h)/h= (h(h+ 3))/h which, as long as h is not 0, is equal to h+ 3. Although we cannot evalate that fraction at x= 0, because both numerator and denominator is 0, we can take the limit as h goes to 0. Since, as long as h is not 0, the difference quotient is h+ 3 so its limit as h goes to 0 is 3. That is why I said, above, the derivative of g at x= 1 is 3.
     
  7. Nov 4, 2013 #6
    wow thanks!
     
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