Question

What is the limit?

Answer 1

`lim_(x->3)(f(x)-f(3))/(x-3)=0`

Explanation:

`f(3)=3^2-6(3)+3=-6`

`f(x)-f(3)=x^2-6x+3-(-6)=x^2-6x+9`

We now have `lim_(x->3)(x^2-6x+9)/(x-3)`

Using L'Hopital's rule, we know that if `f(x)/g(x)=0/0oroo/oo`, then `lim_(x->a)f(x)/g(x)=(f'(a))/(g'(a))`

In this case we have `lim_(x->3)(x^2-6x+9)/(x-3)=(2(3)-6)/(1)=(6-6)/1=0/1=0`