Use algebra to evalute the following limit. lim h → 0 [((3+h)2− 9 )/h]?

Answer 1

(3+h)^2=(3+h)(3+h)
=9+6h+h^2
The equation becomes ((9+6h+h^2)-9)/h
=(6h+h^2)/h
=h(6+h)/h
=6+h
=6+0 when h->0
=6

Answer 2

Hello,

Rewrite as lim h--> 0 [(6 + 2h -9)/h] = (-3 + 2h)/h now write as two fractions lim h--> 0 -3/h + 2h/h so -3/h + 2 as h---> 0 we have 2 as the limit.

Hope This Helps!!