Finding a limit

[bwooten2432]

This is going to seem like a fairly simply problem but the back of the book says a different answer than what I am.

Find the limit.

Lim ( as h approaches 0) ((h-1)^3-1)/h

The book says 3.

So I...
1. expended (h-1)^3
2. noticed that the numbers cancelled.
3. factored out the bottom h.
4. substituted 0 in for h...got O!
5. Not 3.

Help! Thank you.

 

[pka]

Find the limit.
Lim ( as h approaches 0) ((h-1)^3-1)/h
The book says 3.

\(\displaystyle \dfrac{(h-1)^3+1}{h}=h^2-3h+3\)

 

[soroban]

Hello, bwooten2432!

You seem to have a typo . . . It should be \(\displaystyle (h+1)^3\)

\(\displaystyle \displaystyle\text{Find the limit: }\:\lim_{h\to0}\frac{(h+1)^3 - 1}{h}\)

\(\displaystyle \displaystyle \lim_{h\to0}\frac{h^3 + 3h^2 + 3h + 1 - 1}{h} \;=\;\lim_{h\to0}\frac{h^3 + 3h^2 + 3h}{h} \;=\;\lim_{h\to0}\frac{h(h^2+3h + 3)}{h}\)


. . \(\displaystyle \displaystyle =\:\lim_{h\to0}(h^2 + 3h + 3) \:=\:0 + 0 + 3 \:=\:3 \)

 

[bwooten2432]

No, the book is saying (h-1)^ +1.

...maybe the book has a typo.

Thank you so much for the help!

 

[pka]

No, the book is saying (h-1)^ +1.
...maybe the book has a typo.
Thank you so much for the help!

See reply #2