# Limits: h approaching zero

• Dec 13th 2008, 11:53 AM
overduex
Limits: h approaching zero
Can someone please explain how to solve this?

Let f(x)=x^3

What number does [f(5+h)-f(5)]/h approach as h approaches zero?
• Dec 13th 2008, 12:56 PM
thelostchild
We have
$\lim_{h\rightarrow0}\frac{(5+h)^3-5^3}{h}$

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

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

$=3\times5^2=75$
• Dec 13th 2008, 01:06 PM
overduex
Quote:

Originally Posted by thelostchild
We have
$\lim_{h\rightarrow0}\frac{(5+h)^3-5^3}{h}$

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

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

$=3\times5^2=75$

Would you mind explaining how you got: $=\lim_{h\rightarrow0}\frac{5^3+3\times5^2h+3\times 5h^2-5^3}{h}$ ?
I don't quite follow. And where did those 3's come from?
Thank you.
• Dec 13th 2008, 01:09 PM
mr fantastic
Quote:

Originally Posted by overduex
Would you mind explaining how you got: $=\lim_{h\rightarrow0}\frac{5^3+3\times5^2h+3\times 5h^2-5^3}{h}$ ?
I don't quite follow. And where did those 3's come from?
Thank you.

(5 + h)^3 has been expanded. By the way .... One of the terms is missing - it's left for you to find.
• Dec 13th 2008, 01:28 PM
thelostchild
Quote:

Originally Posted by mr fantastic
(5 + h)^3 has been expanded. By the way .... One of the terms is missing - it's left for you to find.

I really should be less asleap while doing it!
• Dec 13th 2008, 01:31 PM
Jhevon
Quote:

Originally Posted by overduex
...I don't quite follow. And where did those 3's come from?
Thank you.

just to add a bit to what Mr F said,

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

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 $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

now plug in your values for a and b
• Dec 13th 2008, 01:38 PM
overduex
Quote:

Originally Posted by mr fantastic
(5 + h)^3 has been expanded. By the way .... One of the terms is missing - it's left for you to find.

Ok, so I expanded it but I still can't get the answer. Can you check for a mistake somewhere?
$=\lim_{h\rightarrow0}\frac{5+h\times 25+10h+h^2- 125}{h}$
$=\lim_{h\rightarrow0}\frac{125+ 50h+5h+25h+ 10h+ 10h^2 + h^3 - 125}{h}$
$=\lim_{h\rightarrow0}\frac{80h +10h^2 +h^3}{h}$
so all I factored out the h and ended up with

$={80 + 10h + h^2}$
which would mean the limit would be approaching 80, but thats not the answer... Help? Please?
• Dec 13th 2008, 01:57 PM
Ziaris
You expanded incorrectly. Use Jhevon's easy way by the binomial theorem to check your expansion. You're supposed to get

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