# [SOLVED] Limits

• Feb 1st 2009, 12:49 AM
>_<SHY_GUY>_<
[SOLVED] Limits
this part of finding the limit [in this form] ALWAYS confuses me, i really can get it [maybe the (h) throws me off]

the limit of [f(x+h)-f(x)] / h
1. f(x) = 3x-1

How does 3x-1 substitute in when there is an "h", it is confusing
• Feb 1st 2009, 12:52 AM
Moo
Hello,
Quote:

Originally Posted by >_<SHY_GUY>_<
this part of finding the limit [in this form] ALWAYS confuses me, i really can get it [maybe the (h) throws me off]

the limit of [f(x+h)-f(x)] / h
1. f(x) = 3x-1

How does 3x-1 substitute in when there is an "h", it is confusing

$f({\color{red}x})=3{\color{red}x}-1$

$f({\color{red}x+h})=3({\color{red}x+h})-1$

So $\frac{f(x+h)-f(x)}{h}=\frac{[3(x+h)-1]-[3x-1]}{h}=\frac{3x+3h-1-3x+1}{h}=\frac{3h}{h}$

Simplify (Nod)
• Feb 1st 2009, 12:53 AM
red_dog
$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{[3(x+h)-1]-(3x-1)}{h}=\lim_{h\to 0}\frac{3x+3h-1-3x+1}{h}=$

$=\lim_{h\to 0}\frac{3h}{h}=3$
• Feb 1st 2009, 12:54 AM
>_<SHY_GUY>_<
Quote:

Originally Posted by Moo
Hello,

$f({\color{red}x})=3{\color{red}x}-1$

$f({\color{red}x+h})=3({\color{red}x+h})-1$

So $\frac{f(x+h)-f(x)}{h}=\frac{[3(x+h)-1]-[3x-1]}{h}=\frac{3x+3h-1-3x+1}{h}=\frac{3h}{h}$

Simplify (Nod)

hmm...right(Happy)...but it says [and i forgot to mention that x approaches zero [the limit] the answer 3?
• Feb 1st 2009, 12:55 AM
>_<SHY_GUY>_<
nevermind....errrrr(Doh) im so dumb...thank you, i get it now haha