Help with this one limit problem

[Jess]

I thought I could figure it out...I got an answer, but apparently it's wrong.

http://i45.tinypic.com/23i6vdz.gif

I got 1/40 but it's wrong (submitting online, so it's telling me it's incorrect) :/ I don't know what else to do or what I did wrong....I simply simplified the denominator by using common denominators ---- 40(40+h) and went on from there, but no matter what I still get 1/40 and I can't see what I did wrong. must be a stupid algebra mistake

[Gigablue]

I got 1/40 as well.

First step. Simplify the 40/40 to 1. Getting lim ((40/40+x)-1)/x

Next. Convert the 1 into (40+x)/(40+x) then combine the fractions, getting lim (40-(40+x)/(40+x))/x

The 40s cancel and give lim (x/(40+x))/x

The Xs cancel and give lim 1/(40+x)

As x goes to 0 the limit goes to 1/40.

I think you got the right answer.

[Jess]

there must be something wrong with that online site then. mk :/

[scer123]

The limit=0

[(40/(40+x))-(40/40)]/x

[(40/(40+x))-1]/x

[(40-(40+x))/(40+x)/x

(40x-40x+x^2)/(40+x)

Lim (0)^2/(40+(0))=0
x->0
------------------------------------------------------------------
Just kidding! It's actually -1/40

Cause (40-40-x)/(40+x)/x

(-x)/(40+x) x 1/x

-1/(40+x)

Please don't double post-Silver Assassin

[Jess]

Yeah, that was the mistake. It was negative, not positive. Anyways, I already got that answer long ago, but thanks anyways