Question

How do you use the limit definition to find the derivative of `f(x)=x^2-15x+7`?

Answer 1

Expand, reduce and evaluate the limit.

Explanation:

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

Long version explanation

For `f(x) = x^2-15x+7`,

notice that `f(x+h) = (x+h)^2-15(x+h)+7`.

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

`= lim_(hrarr0)(((x+h)^2-15(x+h)+7)-(x^2-15x+7))/h`

Notice that, if we try to evaluate by substitutiton, we get the indeterminate form `0/0`.

Expand the numerator:
(note `(x+h)^2 = x^2+2xh+h^2` `" "` use FOIL if you need to)

`= lim_(hrarr0)((x^2+2xh+h^2-15x-15h+7)-(x^2-15x+7))/h`

`= lim_(hrarr0)((x^2+2xh+h^2-15x-15h+7-x^2+15x-7))/h`

Now, some of the terms in the numerator add to `0`

`= lim_(hrarr0)((color(red)(x^2)+2xh+h^2 color(green)(-15x) -15hcolor(blue)(+7) color(red)(-x^2) color(green)(+15x)color(blue)(-7)))/h`

`= lim_(hrarr0)(2xh+h^2-15h)/h`

We still get `0/0`, but we can factor and reduce

`= lim_(hrarr0)(cancel(h)(2x+h-15))/cancel(h)_1`

`= lim_(hrarr0)(2x+h-15)`

And now we can evaluate the limit

`= 2x+(0)-15 = 2x-15`

So, `f'(x) = 2x-15`

Short version

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

`= lim_(hrarr0)(((x+h)^2-15(x+h)+7)-(x^2-15x+7))/h`

`= lim_(hrarr0)((x^2+2xh+h^2-15x-15h+7-x^2+15x-7))/h`

`= lim_(hrarr0)(2xh+h^2-15h)/h`

`= lim_(hrarr0)(2x+h-15)`

`= 2x-15`