Question

Help with limits please

Answer 1

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

Answer 2

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

Answer 3

multiply

Answer 4

Did you try simplifying the numerator...

Answer 5

are you familiar with L'Hospital Rule ?

Answer 6

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

Answer 7

this comes way way before l'hopital

Answer 8

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

Answer 9

\[(2+x)^3=?\]

Answer 10

yeah it is always algebra that messes up

Answer 11

\[x^{3}+6x ^{2}+12x+8\]

Answer 12

is there a trick on multiplying that out faster?

Answer 13

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

Answer 14

\[x ^{2}+6x+12\]

Answer 15

\[(0)^{2}+6(0)+12=12\]

Answer 16

\[(a+b)^3 =a^3+b^3+3a^2b+3ab^2\]
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

Answer 17

oh wow i forgot about that forumla :o tyvm

Answer 18

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

Answer 19

Haha it does :P