Question

lim x=0 (x+2)^3 - 8 / x

Answer 1

is this an l'hopitals rule question?

Answer 2

If you use L'hospital's rule you'll get
\[\lim_{x \rightarrow 0}\frac{3(x+2)^2}{1} \\ \\ \lim_{x \rightarrow 0} 3(x+2)^2\]
Solve

Answer 3

But the numerator doesn't go to 0?

Answer 4

Oh wait, yes it does... derp :3

Answer 5

yes its a indetermination 0/0

Answer 6

Well, @.Sam. pretty much clears it up :D

Answer 7

what was wrong with the answer @.Sam. gave?

Answer 8

@MarilynRobles you factorized a bit wrong, if you're going to magically add 8 to one side, you need to add 8 to the other side. you have 18 instead

Answer 9

thaknss

Answer 10

why 8!! pleasee can you draw? its hard to me understand youu i dont speak english

Answer 11

err sorry.

x^3 + 18x -8 = 0

x^3 + 18x -8 +8 = +8

that was the step with a mistake

Answer 12

and the x in the denominator? equation becomes?

Answer 13

Use L'Hopital's rule to answer this problem, do not factorize

when you plug in 0 and the equation is indeterminate, take the derivative of the numerator and denominator then plug in 0 again

Answer 14

The mistake is the expansion of \((x+2)^3\)
\[x^3+6 x^2+12 x+8\]
\[\lim_{x \rightarrow 0}\frac{x^3+6 x^2+12 x+8-8}{x} \\ \\ \lim_{x \rightarrow 0}\frac{x^3+6 x^2+12 x}{x}\\ \\ \lim_{x \rightarrow 0}\frac{(x)(x^2+6 x+12) }{x} \\ \\ \lim_{x \rightarrow 0}x^2+6 x+12 \]
Substitute x=0,
12

Answer 15

thanks to all :)

Answer 16

http://openstudy.com/study#/updates/51a2fecde4b00f621d792537