Can someone please explain how to solve this?

Thanks in advance.

Let f(x)=x^3

What number does [f(5+h)-f(5)]/h approach as h approaches zero?

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- Dec 13th 2008, 11:53 AMoverduexLimits: h approaching zero
Can someone please explain how to solve this?

Thanks in advance.

Let f(x)=x^3

What number does [f(5+h)-f(5)]/h approach as h approaches zero? - Dec 13th 2008, 12:56 PMthelostchild
We have

$\displaystyle \lim_{h\rightarrow0}\frac{(5+h)^3-5^3}{h}$

$\displaystyle =\lim_{h\rightarrow0}\frac{5^3+3\times5^2h+3\times 5h^2-5^3}{h}$

$\displaystyle =\lim_{h\rightarrow0}{3\times5^2+3\times5h}$

$\displaystyle =3\times5^2=75$ - Dec 13th 2008, 01:06 PMoverduex
- Dec 13th 2008, 01:09 PMmr fantastic
- Dec 13th 2008, 01:28 PMthelostchild
- Dec 13th 2008, 01:31 PMJhevon
just to add a bit to what Mr F said,

the hard way: write as $\displaystyle (5 + h)(5 + h)(5 + h)$

and expand the brackets two at a time. that is, leave the last one and expand the first two. then expand the remaining two brackets

the easy way: note that $\displaystyle (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

now plug in your values for a and b - Dec 13th 2008, 01:38 PMoverduex
Ok, so I expanded it but I still can't get the answer. Can you check for a mistake somewhere?

$\displaystyle =\lim_{h\rightarrow0}\frac{5+h\times 25+10h+h^2- 125}{h}$

$\displaystyle =\lim_{h\rightarrow0}\frac{125+ 50h+5h+25h+ 10h+ 10h^2 + h^3 - 125}{h}$

$\displaystyle =\lim_{h\rightarrow0}\frac{80h +10h^2 +h^3}{h}$

so all I factored out the h and ended up with

$\displaystyle ={80 + 10h + h^2}$

which would mean the limit would be approaching 80, but thats not the answer... Help? Please? - Dec 13th 2008, 01:57 PMZiaris
You expanded incorrectly. Use Jhevon's easy way by the binomial theorem to check your expansion. You're supposed to get

$\displaystyle \lim_{h\to 0}\frac{5^3 + 75h+15h^2+h^3 - 5^3}{h}.$