Question

Using the limit definition, how do you differentiate `f(x) = x^2 - 1598`?

Answer 1

Expand and simplify. Then evaluate the limit.

Explanation:

`f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h`

For this function, we get

`f'(x) = lim_(hrarr0)([(x+h)^2-1598] - [x^2-1598])/h`

If we try to evaluate the limit by substitution, we get the indeterminate form `0/0`, so we need a different approach.
It may not be clear that expanding the numerator will help, but we need to try something, so let's do it and see what happens.

`f'(x) = lim_(hrarr0)([x^2+2xh+h^2-1598] - [x^2-1598])/h`

`= lim_(hrarr0)(color(red)(x^2)+2xh+h^2color(blue)(-1598) color(red)(-x^2)color(blue)(+1598))/h`

`= lim_(hrarr0)(2xh+h^2)/h`

If we try substitution, we still get indeterminate form `0/0`.

Note however, that for all `h` other that `0`, we have

`(2xh+h^2)/h = (cancel(h)(2x+h))/cancel(h) = 2x+h`

Therefore,

`f'(x) = lim_(hrarr0)(2xh+h^2)/h = lim_(hrarr0)(2x+h)=2x`

In many classes it is permissible to write this without comment as follows:

`f'(x) = lim_(hrarr0)([(x+h)^2-1598] - [x^2-1598])/h`

`= lim_(hrarr0)([x^2+2xh+h^2-1598] - [x^2-1598])/h`

`= lim_(hrarr0)(x^2+2xh+h^2-1598 -x^2+1598)/h`

`= lim_(hrarr0)(2xh+h^2)/h`

`= lim_(hrarr0)(2x+h)`

`= 2x`