Question

Limit
http://puu.sh/d2ZeU/074c93dcf0.png

Answer 1

infinity/..

Answer 2

nope

Answer 3

hmmm tried expanding the binomial yet?

Answer 4

when i expand and then apply 0 to h it's undefined right?

Answer 5

wel... what did you get from the binomial expansion though?

Answer 6

\(\bf lim_{h\to 0}\ \cfrac{(2+h)^3-8}{h}
\\ \quad \\
\cfrac{(2+h)^3-8}{h}\implies \cfrac{\square? +\square? h+\square? h^2+h^3-8}{h}?\)

Answer 7

u can use the formuala\((a^3-b^3)\)

\(\Large (2+h)^3-2^3\)

Answer 8

but if zero is on the bottom inst it undefined?

Answer 9

yes.... " IF " zero is at the bottom, IF

Answer 10

so... what does the binomial expansion give you anyway?

Answer 11

left side approaches 0, positive infinity, right side approaches 0, negative infinity

Answer 12

wait no..

Answer 13

But you will see that once you expand this limit then divide each of the values in the numerator by "h", then the function will simplify whereas some values will approach 0 and some will not. :)

Answer 14

So first expand \((2+h)^3\)

Answer 15

i expanded and got 8+12h+6h^2 + h^3

Answer 16

12h?

Answer 17

ahemm \(\bf lim_{h\to 0}\ \cfrac{(2+h)^3-8}{h}
\\ \quad \\
\cfrac{(2+h)^3-8}{h}\implies \cfrac{\cancel{ 8 } +6 h+6 h^2+h^3\cancel{ -8 }}{h}\implies \cfrac{\cancel{ h }(6+6h+h^2)}{\cancel{ h }}\)

so... what do you think is the limit?

Answer 18

hmmm wait a sec... I did expaded.. off =) you're correct.. should be 12.. anyhow, lemme rewrite that

Answer 19

\(\bf lim_{h\to 0}\ \cfrac{(2+h)^3-8}{h}
\\ \quad \\
\cfrac{(2+h)^3-8}{h}\implies \cfrac{\cancel{ 8 } +12 h+6 h^2+h^3\cancel{ -8 }}{h}\implies \cfrac{\cancel{ h }(12+6h+h^2)}{\cancel{ h }}\)

and surely you could see the limit

Answer 20

oh tyvm

Answer 21

yw

Answer 22

Yeah, \((2+h)^3 = 2^3 + 3 \cdot 2\cdot 2 \cdot h + 3\cdot 2 \cdot h^2 + h^3\) well, thats how I remember how to expand it. Lol.

Answer 23

The numbers decrease as the h's increase. Woo!

Answer 24

oh thank you for the tip