f^1(a)=lim as x goes to 0,, 1/(a+h)^2- 1/(a)^2 /h

## find the lim as x goes to 0

# 2 Answers

In this problem, you are actually being asked to derive the derivative of f(x) = 1/x^2 (i.e., 1 over x squared).

You are doing this using the formal definition of the derivative, although in algebra class (or precalculus), this is called the "difference quotient". The general form of the difference quotient for a function f(x) is the following:

DQ = [ f(x+h) - f(x) ] / h.

(You can think of this as the average slope of the graph of f(x) between the points x and x+h.)

If we let h go to zero, we say this quotient becomes the "derivative"... in other words, the difference quotient becomes the derivative of f(x) at the limit when h goes to 0. (The derivative is the "instantaneous" slope of f(x).)

lim DQ = lim { [ f(x+h) - f(x) ] / h } = f'(x) = df/dx

(f'(x) and df/dx are two different notations for the derivative of f(x). )

Now let's look at the problem you were asked to do:

In this problem f(x) = 1/x^2. Taking the limit of the difference quotient using this f(x) yields:

lim { [ f(x+h) - f(x) ] / h } = lim { [ 1/(x+h)^2 - 1/x^2 ] / h }

The "difficult" part of this problem for most of us is doing the algebra to simplify this expression, but here goes:

(1) The numerator [ 1/(x+h)^2 - 1/x^2 ] can be combined into a single "fraction" if we create the common denominator:

(x^2)(x+h)^2 (i.e. multiply the x^2 by the (x+h)^2)

Doing this gives us:

[ 1/(x+h)^2 - 1/x^2 ] / h = [x^2 - (x+h)^2] / [h (x^2) (x+h)^2]

(2) Expanding the binomial terms (i.e,, FOILing the two (x+h)^2 terms):

= [x^2 - x^2 - 2xh - h^2] / [h (x^2) (x^2 + 2xh + h^2)]

= (-2xh - h^2) / [ (x^4h + 2x^3h^2 + x^2h^3) ]

(3) Now divide both the numerator and the denominator by h:

= (-2x - h) / [(x^4 + 2x^3h + x^2h^2)]

Whew! (I told you the algebra was messy!) Finally,we can take the limit (i.e., let h go to 0), which means that any term that is multiplied by h (or h^2, etc.) will go away (become 0):

lim { (-2x - h) / [(x^4 + 2x^3h + x^2h^2)] } = -2x / x^4 = -2/x^3

(i.e., negative 2 over x cubed). And there's your answer.

This is exactly the derivative of f(x) = 1/x^2. Once you learn calculus, this problem becomes a very simple application of the "power rule" and can be done in one step!

Power rule: If f(x) = x^n then f'(x) = n x^(n-1) (for all n other than 0).

so if f(x) = 1/x^2 = x^(-2) then f'(x) = (-2) x^(-2 - 1) = -2 x^(-3) = -2/x^3 .

There's no x in your function. I take it you mean as H goes to 0? That's generally how you use the difference quotient for a particular function. Does it make more sense that way, then?