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

[Arixal]

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.

 

[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 $h$.

 

[DonAntonio]

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