What is the limit of: (3x²+2x-1)/(9x²-1) when x->1/3?
The answer is 2/3 but it doesn't the same result i did
The answer is 2/3 but it doesn't the same result i did
-
Start out by factoring the numerator and denominator to get:
(a) 3x^2 + 2x - 1 = (3x - 1)(x + 1)
(b) 9x^2 - 1 = (3x + 1)(3x - 1).
So, we have:
lim (x-->1/3) (3x^2 + 2x - 1)/(9x^2 - 1)
= lim (x-->1/3) [(3x - 1)(x + 1)]/[(3x + 1)(3x - 1)], by factoring
= lim (x-->1/3) (x + 1)/(3x + 1), by canceling 3x - 1
= (1/3 + 1)/(1 + 1)
= 2/3, as required.
I hope this helps!
(a) 3x^2 + 2x - 1 = (3x - 1)(x + 1)
(b) 9x^2 - 1 = (3x + 1)(3x - 1).
So, we have:
lim (x-->1/3) (3x^2 + 2x - 1)/(9x^2 - 1)
= lim (x-->1/3) [(3x - 1)(x + 1)]/[(3x + 1)(3x - 1)], by factoring
= lim (x-->1/3) (x + 1)/(3x + 1), by canceling 3x - 1
= (1/3 + 1)/(1 + 1)
= 2/3, as required.
I hope this helps!
-
rewrite: (x+1)/(3 x+1)
plug in x=1/3
answer=2/3
plug in x=1/3
answer=2/3
-
lim x->1/3 (3x² + 2x - 1)/(9x² - 1)
= lim x->1/3 (3x - 1)(x + 1)/(3x + 1)(3x - 1) = lim x->1/3 (x + 1)/(3x + 1) = (4/3)/(2) = 2/3
= lim x->1/3 (3x - 1)(x + 1)/(3x + 1)(3x - 1) = lim x->1/3 (x + 1)/(3x + 1) = (4/3)/(2) = 2/3
-
Hmm...it's equal zero so we have to resolve by factoring....