Question

How do you find the limit of `(x^2-4x+23)` as x approaches 2?

Answer 1

`lim_(x->2) (x^2-4x+23) = 19`

Explanation:

`f(x) = (x^2-4x+23)` is a polynomial and as such it is continuous for every `x in RR`

By definition for a continuous function:

`lim_(x->a) f(x) = f(a)`

So:

`lim_(x->2) (x^2-4x+23) = [x^2-4x+23]_(x=a) = (4-8+23) =19`