Question

whats the limit of the following function? could somebody help me please?

Answer 1

\[\lim \frac{ 2x ^{2} + x - 1}{x ^{3} -x}\]

Answer 2

x--> -1

Answer 3

0

Answer 4

just plug in -1

Answer 5

0/2

Answer 6

Use L'Hopital rule to get it.

Answer 7

what is that L rule? sorry i dont know much about this :(

Answer 8

looks factorable, no?

Answer 9

hmm i got 1 with L'hopitals

Answer 10

yeah it's factorable, no l'hospital required

Answer 11

\[\frac{(2x-1)(x+1)}{x(x-1)(x+1)}\]

Answer 12

o right there is no indeterminate so no need to use L'hopitals

Answer 13

yeep i did that already then i get:\[\frac{( 2x-1)(x+1) )}{ x(x+1)(x-1) }\] but then i get stuck

Answer 14

cancel x+1...

Answer 15

\[\frac{(2x-1)\cancel{(x+1)}}{x(x-1)\cancel{(x+1)}}\]

Answer 16

oh thats an x^3 in the denominator

Answer 17

so i'll have: \[\frac{ 2(x-1) }{ (x-1)}\] and i can cancel (x-1) right?

Answer 18

can't factor 2 out like that

Answer 19

how did 2 move outside the parentheses, and where did the other x in the denominator go?

Answer 20

\[\frac{(2x-1)\cancel{(x+1)}}{x(x-1)\cancel{(x+1)}}=\frac{2x-1}{x(x-1)}\]now you can just plug in

Answer 21

ooh thank you so much! so the the limit the is equal to:-3/2 right?

Answer 22

yes :)

Answer 23

thank you so much !!