Question

limit x->1 x^3 -3x +2 / x-1

Answer 1

I get 0/0 and o supose its wrong :(

Answer 2

do you know lhospital's law?

Answer 3

I have read about it I dont know how to use it

Answer 4

you need to build the derivats of the two functions

Answer 5

you take the derivative of the nominator and the denominator
(x^3-3x+2)' = ?
(x-1)' = ?

Answer 6

If i set 1 in for x I get 1^3 - 3 (1) +2 / 1-1 and get 0/0 = 0

Answer 7

limit is x to 1

Answer 8

(x^3-3x+2)' = 3x^2 -3
(x-1)' = 1
now :
lim x->1 of (3x^2 - 3) / 1 = 0

Answer 9

\[\lim_{x \rightarrow 1}\] x^3 - 3x +2 / x -1

Answer 10

okay so It does not have to be 1? since the limit goes to 1

Answer 11

it goes to 0 as coolsector wrote

Answer 12

http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

Answer 13

if you dont know lhospitals
we can do as well :
nominator : x^3 -x -2x -2
x(x^2 - 1) -2(x-1)
x(x-1)(x+1) -2(x-1)

so the limit :
(x-1)(x(x+1)-2) / (x-1)
->
lim of x(x+1)-2 as x goes to 1 = 0.

Answer 14

this is another way if you dont know lhospital's..

Answer 15

Okay Thanks :)

Answer 16

yw

Answer 17

by l'Hospital's rule differentiating the numerator and denominator... we get Lim x->1 3x^2-3/1 applying limit we can get 3(1)^2 -3 = 3-3 =0