2007-01-24 17:44:55 · 4 answers · asked by guy r 2 in Science & Mathematics ➔ Mathematics
You solve this the same way as the other question you posted.
When you direct substitute, you get an indeterminate form 0/0 so use L'Hospitals' Rule
take the derivative of the top and bottom:
[-7/(x+5)^2]/1
now direct substitute:
[-7/(0+5)^2]/1 = -7/25
2007-01-24 17:58:26 · answer #1 · answered by Anonymous · 0⤊ 0⤋
lim xâ0 of (-7/5 + 7/(x + 5))/x
= lim xâ0 of 7(-1/5 + 1/(x + 5))/x
= lim xâ0 of 7{(-x - 5 + 5)/[5(x + 5)]}/x
= lim xâ0 of 7{-x/[5(x + 5)]}/x
= lim xâ0 of -7x/[5x(x + 5)]
= lim xâ0 of -7/[5(x + 5)] = -7/25
2007-01-24 18:20:59 · answer #2 · answered by Northstar 7 · 0⤊ 0⤋
(-7 / 5 + 7 / (x + 5)) / x =
7(x / (x + 5) - x / 5)
lim(7(x / (x + 5) - x / 5)) =
x->0
lim(7(1 / 5 - x / 5)) = 7/5
x->0
2007-01-24 17:54:58 · answer #3 · answered by Helmut 7 · 0⤊ 1⤋
-0.28
x=0-, f(x) -> -0.28
x=0+, f(x) -> -0.28
2007-01-24 18:01:35 · answer #4 · answered by noobie 2 · 0⤊ 0⤋