# Difference between a limit and a derivative?

1. Nov 4, 2013

### AakashPandita

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. Nov 4, 2013

### Nick O

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

3. Nov 4, 2013

### 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.

4. Nov 4, 2013

### Nick O

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.

5. Nov 4, 2013

### HallsofIvy

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: $\lim_{x\to 0} g(x)= -1$.

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.

6. Nov 4, 2013

wow thanks!