- #1

- 9

- 0

^{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 are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Arixal
- Start date

- #1

- 9

- 0

- #2

pwsnafu

Science Advisor

- 1,080

- 85

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

- #3

- 606

- 1

^{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 [itex]\,h\to 0\,[/itex] 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

Share: