Question

Using the limit definition, how do you find the derivative of `3/(x-2)`?

Answer 1

Given `f(x) = 3/(x-2)`, then `f(x+h) = 3/(x+h-2)`

The limit definition is:

`f'(x) = lim_(h to 0)(f(x+h)-f(x))/h`

Substitute the functions into the definition:

`f'(x) = lim_(h to 0)(3/(x+h-2)-3/(x-2))/h`

Eliminate the fractions in the numerator by multiplying by 1 in the form `((x-2)(x+h-2))/((x-2)(x+h-2))`:

`f'(x) = lim_(h to 0)(3/(x+h-2)-3/(x-2))/h((x-2)(x+h-2))/((x-2)(x+h-2))`

Please observe that we have not changed the value of the limit, because we only multiplied by one in a special form.

Perform the multiplications:

`f'(x) = lim_(h to 0)((3(x-2)(x+h-2))/(x+h-2)-(3(x-2)(x+h-2))/(x-2))/(h(x-2)(x+h-2))`

Please observe how the factors in the numerator cancel to eliminate the fractions:

`f'(x) = lim_(h to 0)((3(x-2)cancel(" "(x+h-2)))/cancel(" "(x+h-2))-(3cancel(" "(x-2))(x+h-2))/cancel(" "(x-2)))/(h(x-2)(x+h-2))`

Leaving us with the following:

`f'(x) = lim_(h to 0)(3(x-2)-3(x+h-2))/(h(x-2)(x+h-2))`

Use the distributive property to eliminate the parenthesis in the numerator:

`f'(x) = lim_(h to 0)(3x-6-3x-3h+6)/(h(x-2)(x+h-2))`

Combine like terms in the numerator:

`f'(x) = lim_(h to 0)(-3h)/(h(x-2)(x+h-2))`

`h/h` cancels:

`f'(x) = lim_(h to 0)(-3)/((x-2)(x+h-2))`

Let `h to 0`:

`f'(x) = (-3)/((x-2)(x-2))`

Write the denominator as a square:

`f'(x) = (-3)/(x-2)^2`