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

  • Thread starter Arixal
  • Start date
  • #1
9
0

Main Question or Discussion Point

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

  • #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 manually calculate the expression for each ##h##.
 
  • #3
606
1
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 [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
 

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

Replies
5
Views
2K
Replies
5
Views
2K
Replies
3
Views
1K
Replies
26
Views
4K
Replies
17
Views
3K
Replies
3
Views
2K
Replies
4
Views
8K
Replies
7
Views
2K
  • Last Post
Replies
14
Views
26K
Replies
4
Views
2K
Top