# Thread: Limits: h approaching zero

1. ## 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?

2. We have
$\lim_{h\rightarrow0}\frac{(5+h)^3-5^3}{h}$

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

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

$=3\times5^2=75$

3. 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.

4. 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.

5. 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!

6. 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

7. 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?

8. You expanded incorrectly. Use Jhevon's easy way by the binomial theorem to check your expansion. You're supposed to get

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