lim x > 0 = (x^2 - x + sinx) / (2x)
note: x^2 is x squared
2007-10-07 15:28:02 · 5 answers · asked by NEEDs MATH HELP!! 1 in Science & Mathematics ➔ Mathematics
Easiest way to do this is to just break it into three limits:
[x→0]lim (x² - x + sin x)/(2x)
1/2 [x→0]lim (x²/x - x/x + sin x/x)
1/2 [x→0]lim (x - 1 + sin x/x)
1/2 (0 - 1 + 1)
1/2 (0)
0
2007-10-07 15:33:44 · answer #1 · answered by Pascal 7 · 0⤊ 0⤋
This easy mate :
lim x > 0 = (x^2 - x + sinx) / (2x)
(x^2 - x + sinx) / (2x)
d/dx = (2x-1+cosx) / 2 = (0-1+1)/2 = 0
2007-10-07 22:34:56 · answer #2 · answered by 1st 2 · 0⤊ 0⤋
Thanks for the parentheses!
lim x-> 0 ... (2x - 1 + cos x) / 2
2(0) - 1 + cos (0) / 2 = (0 -1 + 1) / 2 = 0/2
Answer: 0
2007-10-07 22:33:23 · answer #3 · answered by UnknownD 6 · 0⤊ 0⤋
use L'Hopital's rule:
lim x--> 0 = [2x -1 -cosx ]/2
plug in 0:
=[2(0) -1 cos (0) ]/2
= -1/2
2007-10-07 22:31:43 · answer #4 · answered by sayamiam 6 · 0⤊ 1⤋
you just plug in the zero where there is an "x"
((0^2 - 0 + sin(0)) / (2x0)
2007-10-07 22:32:26 · answer #5 · answered by yahooooo 1 · 0⤊ 0⤋