is it 0?
cause i/i^2 = 0?
2007-02-12 05:30:03 · 5 answers · asked by changchih 7 in Science & Mathematics ➔ Mathematics
lim (i +1) / (i^2 + 3)
n -> infinite
2007-02-12 05:35:03 · update #1
as i approaches inifinity
2007-02-12 05:41:19 · update #2
Then you're right. It's zero, because the i term is "insignificant" compared to the i² term, so it's like taking the lim (i→0) of (1/i)
2007-02-12 05:44:24 · answer #1 · answered by bequalming 5 · 0⤊ 0⤋
Algebraically, divide numerator and denominator by i^2
r = [1/i + 1/i^2]/[1+3/i^2]
now just take i-> inf
r -> 0/1 = 0
2007-02-12 05:54:09 · answer #2 · answered by modulo_function 7 · 0⤊ 0⤋
The problem is, for a limit, i needs to be approaching a value.
Can you add this detail in your question?
2007-02-12 05:33:53 · answer #3 · answered by Puggy 7 · 1⤊ 0⤋
i² = -1
therefore (i + 1) / (i² + 3) = (i + 1) ( -1 + 3) = (i + 1) / 2
But unsure about what question is asking?
2007-02-12 05:49:37 · answer #4 · answered by Como 7 · 0⤊ 0⤋
Limit as i approaches what??!?
2007-02-12 05:33:25 · answer #5 · answered by Anonymous · 1⤊ 0⤋