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

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Answer 1 · 2017-02-02T21:15:26

I got $2$

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

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