Help with limits please

Question

zappy620 · Answer

$$\lim_{x \rightarrow 0}\frac{ (2+x)^{3}-8 }{ x }$$

zappy620 · Answer

The answer in the book is 12 and I can't seem to figure out how they got it.

satellite73 · Answer

multiply

agent0smith · Answer

Did you try simplifying the numerator...

s4sensitiveandshy · Answer

are you familiar with  L'Hospital Rule ?

agent0smith · Answer

^You could but no need for it. Simplify and you'll see why.

satellite73 · Answer

this comes way way before l'hopital

zappy620 · Answer

actually i just figured out i was just multiplying the factors wrong smh

satellite73 · Answer

$$(2+x)^3=?$$

satellite73 · Answer

yeah it is always algebra that messes up

zappy620 · Answer

$$x^{3}+6x ^{2}+12x+8$$

zappy620 · Answer

is there a trick on multiplying that out faster?

zappy620 · Answer

well just to finish off the problem I guess, after factoring out the x's and getting rid of the 8s, im left with

zappy620 · Answer

$$x ^{2}+6x+12$$

zappy620 · Answer

$$(0)^{2}+6(0)+12=12$$

s4sensitiveandshy · Answer

$$(a+b)^3 =a^3+b^3+3a^2b+3ab^2$$
but i guess it's easy to draw a box and distribute (a+B)(a+b) and then quadratic by (a+b)
  helps me to stay way from mistakes

zappy620 · Answer

oh wow i forgot about that forumla :o tyvm

agent0smith · Answer

"Simplify and you'll see why."
That rhymed.

zappy620 · Answer

Haha it does :P