Question

trying to solve lim as x approaches 0 of (h-1)^3+1/h

Answer 1

\[\lim_{h\rightarrow0}\frac{(h-1)^3+1}{h}\]
?

Answer 2

What do you get if you multiply out \((h-1)(h-1)(h-1)\) and then add 1?

Answer 3

h^3-2h^2+1h^2/-h^3+2h^2-1h^2 =1+1 =2
?

Answer 4

No, might need to practice that multiplication a bit...
\[(h-1)(h-1)(h-1) = (h-1)(h^2-1h-1h+1) =\]\[ (h^3-1h^2-1h^2+1h-1h^2+1h+1h-1)=h^3-3h^2+3h-1\]

Answer 5

oops-- then if the lim approaches 0 then replacing h with 0 would give me -1

Answer 6

No again.

\[\lim_{h\rightarrow0}\frac{(h-1)^3+1}{h} = \lim_{h\rightarrow0}\frac{h^3-3h^2+3h-1+1}{h} \]

Answer 7

so another try- can i factor out h as in h(h^2-3h+3)/h am i on the right track?

Answer 8

Exactly!

Answer 9

Thanks!!!!

Answer 10

What was your final answer?

Answer 11

Hopefully, the 3 rd time is the charm ;-)

Answer 12

3 I think thats it

Answer 13

Yes, that's why I wrote my last message the way I did ;-) 3 is correct.