... lim (x^3 - 1)/(x - 1) =
x=>1
I believe it to be 6 but I'm not 100% sure
2007-01-27 03:34:10 · 5 answers · asked by Olivia 4 in Science & Mathematics ➔ Mathematics
I'm not exactly sure how you got six....i get infinity
2007-01-27 03:42:18 · answer #1 · answered by Paulien 5 · 0⤊ 0⤋
Factorise x^3 -1 as (x -1)(x^2 + x + 1), getting
lim (x^3 - 1)/(x - 1)
x -> 1
= lim [(x - 1)(x^2 + x + 1)]/(x - 1)
x -> 1
= lim (x^2 + x + 1)
x -> 1
= 3
Alternately, use L' Hospital's rule to get
lim 3x^2 = 3
x -> 1
2007-01-27 11:42:20 · answer #2 · answered by Avi 2 · 0⤊ 1⤋
... lim (x^3 - 1)/(x - 1) =
x=>1 Sice this is 0/0 it si indeterminate and we must investigate further using L'Hospital's rule. Separately take derivative of numerator and denominator getting:
... lim (x^3 - 1)/(x - 1) =
x=>1
... lim (3x^2)/( 1) = 6.
x=>1
yes, you're correct, the limit is 6.
2007-01-27 11:50:28 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
(x^3-1)/(x-1) = x^2+x+1 so
lim (x^2+x+1) = (1+1+1) = 3
x=>1
so the answer is 3, not 6.
2007-01-27 11:45:53 · answer #4 · answered by SHIRLEY L 2 · 0⤊ 0⤋
if you wrote the funtion as (x^3-1)(x-1)=
(1-1)(1-1)= (0)(0)=0
2007-01-27 11:42:37 · answer #5 · answered by Zidane 3 · 0⤊ 0⤋