# Limit of Lim h->0 ((2+h^2) - 8)/h

I saw my professor solve for the limit of ((2+h)2 - 8)/h as h approaches 0 and it was defined, but for the life of me I can't remember how. Everyone I have asked can't seem to figure it out either, so I was hoping someone here could explain.

## Answers and Replies

pwsnafu
I saw my professor solve for the limit of ((2+h)2 - 8)/h as h approaches 0

I'm going to assume the thread title is a typo.

and it was defined

Hint: let ##h = 0.1, 0.01, 0.001, \ldots## and manually calculate the expression for each ##h##.

I saw my professor solve for the limit of ((2+h)2 - 8)/h as h approaches 0 and it was defined, but for the life of me I can't remember how. Everyone I have asked can't seem to figure it out either, so I was hoping someone here could explain.

You wrote

$$\frac{(2+h)^2-8}{h}$$

and the limit of this thing when $\,h\to 0\,$ doesn't exist. But perhaps you meant

$$\lim_{h\to 0}\frac{(2+h)^3-8}{h}=\lim_{h\to 0}\frac{8+12h+6h^2+h^3-8}{h}=\lim_{h\to 0}\;(12+6h+h^2)=12$$

DonAntonio