Posted by **Leanna** on Tuesday, September 21, 2010 at 6:51pm.

If the function f is defined by f(x)=x^2-2x, find limit h->0

(f(3+h)-(f(3-h))/2h

- Calculus -
**bobpursley**, Tuesday, September 21, 2010 at 7:05pm
first, f(3+h)=(3+h)^2 -2(3+h)

= 9+6h+h^2-6-2h = h^2+4h+3= (h+3)(h+1)

second f(3-h)=(3-h)^2 -2(3-h)=

= 9-6h+h^2-6+2h=h^2-4h+3=(h-3)(h-1)

so first -second then is

h^2-h^2+4h+4h +3-3=8h

finally, lim 8h/2h=4

check all that.

- Calculus -
**Leanna**, Tuesday, September 21, 2010 at 7:18pm
Thank you for your help!

For the second part I thought it would be -(9+6h-h^2)-6+2h?

- Calculus -
**bobpursley**, Tuesday, September 21, 2010 at 8:29pm
you put a minus sign in front of the first term? Why?

- Calculus -
**bobpursley**, Tuesday, September 21, 2010 at 8:30pm
and, the 6h term in (3-h)^2=9-6h+h^2

- Calculus -
**Leanna**, Tuesday, September 21, 2010 at 9:01pm
(f(3+h)-(f(3-h))/2h

I put a minus sign because it has a negative in front of the f(3-h) so it's -(f(3-h))

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