Question

solve the limit; lim x'2 -4/x'3 -8 x→-2

Answer 1

x'?

Answer 2

X^

Answer 3

power

Answer 4

oh like x^2
x^3

Answer 5

yeah, i dont know how to put that with the keyboard

Answer 6

\[\lim_{x \rightarrow -2}\frac{x^2-4}{x^3-8}=\lim_{x \rightarrow -2}\frac{(x-2)(x+2)}{x^3-8}=\lim_{x \rightarrow -2}\frac{(x-2)(x+2)}{(x-2)(x^2+2x+4)}\]

Answer 7

the x-2's cancel

Answer 8

i have x+2/x'2+2x+4, but i dont know what to do next

Answer 9

\[\lim_{x \rightarrow -2}\frac{x+2}{x^2+2x+4}=\frac{-2+2}{(-2)^2+2(-2)+4}=0\]

Answer 10

yes that's it, but is indeterminate

Answer 11

how is it indeterminate

Answer 12

no

Answer 13

we dont have 0/0

Answer 14

he canceled out

Answer 15

she*

Answer 16

we have 0/(4-4+4)=0/4=0

Answer 17

oh i see, then is correct, he*

Answer 18

she* canceled out stuff so that it was no longer indeterminate form when youget an indeterminate form in calculus 1, the only way ou can solve it is by manipulation of the problem

Answer 19

we had 0/0 at the begining but like out said i canceled the x-2's so no more 0/0

Answer 20

whether that be cancelling factors or rationalizing

Answer 21

also do you know l'hospital
if you know way i can show you another way to find the limit here

Answer 22

ok, thanks, i was confused

Answer 23

please, show me how

Answer 24

oops wait don't look at that

Answer 25

oh, a derivative

Answer 26

wiat what's happening
what i'm doing

Answer 27

didn't i get 0 above?

Answer 28

is 0/12?

Answer 29

\[\lim_{x \rightarrow -2}\frac{x^2-4}{x^3-8}=\lim_{x \rightarrow -2}\frac{2x}{3x^2}=\lim_{x \rightarrow -2}\frac{2}{3x}=\frac{2}{3(-2)}=\frac{1}{-3}\]
ok something is wrong with one of my ways

Answer 30

let me look above

Answer 31

omg out what's wrong with what i did?

Answer 32

out are you still there?

Answer 33

let me get zarkon

Answer 34

i don't get it, but you were right at the top

Answer 35

ok i did it two ways
and i didn't get the samething both ways
so either im confused about calculus or algebra

Answer 36

(-2)^3 = -8 for starters

Answer 37

omg no way

Answer 38

you can't use L'Hospitals unless you have an indeterminant form

Answer 39

i suck at algebra

Answer 40

look like we do

Answer 41

i thought i had 0/0
mu we didn't have 0/0
lol

Answer 42

except the second time you use l'hospitals rule

Answer 43

so we can't use l'hospital way sorry

Answer 44

i used l'hospital once

Answer 45

lHopital is effective on indeterminant forms :)

Answer 46

lol..nope you don't

Answer 47

\[\lim_{x->-2}\frac{x^2-4}{x^3-8}=\frac{(-2)^2-4}{(-2)^3-8}=\frac{0}{-16}=0\]

Answer 48

at the beginning