[f(x+h) - f(x)]/h Help!

**snypeshow** · October 28th 2010, 08:00 AM

Hey guys, my math prof gave me a problem i'm not sure exactly how to solve.

f(x) = (x+1)^2
lim h->0

I'm guessing the pattern is: [f(x+h) - f(x)]/h

so you would get this: [(x+1+h)^2 - (x+1)^2]/h

and the next step is: [(x+1)^2 + 2(x+1)h + h^2 - (x+1)^2]/h

i dont get how to go from the first step to the second step =/

(i tried to use the math tags but they didnt come out right, sorry for not formatting them correctly)

**Plato** · October 28th 2010, 08:07 AM

Here is the way LaTeX works at this site: [tex]\dfrac{f(x+h)-f(x)}{h} [/tex] gives $\dfrac{f(x+h)-f(x)}{h}$

I suggest using $f(x)=x^2+2x+1$

**snypeshow** · October 28th 2010, 08:44 AM

i keep trying to use it but i get a syntax error. Here's a picture of what i dont understand. How do you get from the first line to the second line?

**harish21** · October 28th 2010, 08:56 AM

$(x+1+h)^2$

Consider this term as $(a+b)^2$ ; where a = x+1 and b = h

$(a+b)^2 = a^2+2ab+b^2$

so,

$(x+1+h)^2=(x+1)^2+2(x+1)h+h^2$

**snypeshow** · October 28th 2010, 09:08 AM

thanks! i understand how they got from the second to the third line, but how do you get from the third line to the fourth line?

**harish21** · October 28th 2010, 09:27 AM

$\dfrac{2(x+1)h+h^2}{h} = \dfrac{h[2(x+1)+h]}{h}$

the h on the numerator and the h on the denominator cancel each other out and you are left with

$\displaystyle{\lim_{h \to 0} (2(x+1)+h)}$

**Plato** · October 28th 2010, 09:34 AM

Using ‘\displaystyle’ makes the limit display correctly.
[tex] \displaystyle\lim_{h \to 0} (2(x+1)+h) [/tex] gives $\displaystyle\lim_{h \to 0} (2(x+1)+h)$

**harish21** · October 28th 2010, 09:40 AM

Originally Posted by Plato

Using ‘\displaystyle’ makes the limit display correctly.
[tex] \displaystyle\lim_{h \to 0} (2(x+1)+h) [/tex] gives $\displaystyle\lim_{h \to 0} (2(x+1)+h)$

Did the Editing.

Thanks for the input.

**snypeshow** · October 28th 2010, 11:17 AM

thanks guys! this forum is such a great help

**TheCoffeeMachine** · October 28th 2010, 12:50 PM

Here is how I always do it:

$\displaystyle f'(x) = \lim_{h\to{0}}\frac{(1+x+h)^2-(1+x)^2}{h}$

Let $1+x+h = a$ and $1+x = b$ ,

then $h = a-(1+x) = a-b$ .

Also $h\to{0} \Leftrightarrow {a-b}\to{0} \Rightarrow a\to{b}$

Thus $\displaystyle f'(x) = \lim_{a\to{b}}\frac{a^2-b^2}{a-b} = \lim_{a\to{b}}\frac{(a-b)(a+b)}{a-b}[/Math]<br /> <br /> [Math]\Rightarrow \displaystyle f'(x) = \lim_{a\to{b}}(a+b) = 2b = 2(1+x)$ .

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