Question

How do you evaluate the limit `-x^3+10x^2-33x+38` as x approaches `3`?

Answer 1

I got `2`

Explanation:

In this case we can substitute directly our value to find the limit:
`lim_(x->3)(-x^3+10x^2-33x+38)=-(3)^3+10(3)^2-33*2+38=-27+90-99+38=2`