How to use the limit definition to find....

f[prime](3) for f(x)=x^2-3x-5

Thank You

(f(x + h) - f(x)) / h
= (((x + h)^2 - 3(x + h) - 5) - (x^2 - 3x - 5)) / h
= (x^2 + 2hx + h^2 - 3x - 3h - 5 - x^2 + 3x + 5) / h
= (2hx + h^2 - 3h) / h
= 2x + h - 3
If h approaches zero we have:
= 2x - 3
= f'(x)
Therefore, f'(3) = 2*3 - 3 = 6 - 3 = 3

fprime = (f(x+h)-f(x))/h ** This is the definition of a limit where h is an infinitesimally small number**

f(x+h) = (x+h)^2 -3(x+h) - 5 = x^2 + 2xh + h^2 - 3x - 3h - 5
f(x) = x^2 - 3x - 5

f(x+h) - f(x) = (2xh + h^2 - 3h)

(f(x+h) - f(x))/h = 2x - 3 + h (however h is infinitesimally small so we assume that it is negligible compared to 2x and -3)

f'(3) = 2*3 - 3 = 3

f '(x) = limit h->0 [(f(x+h) - f(x))/h]

f '(3) = limit h->0 [(((3+h)^2 - 3(3+h) - 5) - (3^2 - 3*3 - 5))/h]

= limit h->0 [(h^2 + 3h)/h]
= limit h->0 [h + 3]
= 3