limit x intends to 0(sinx+i)divide by(sinx-1)?

0 ⤊

2007-09-07 02:10:40 · 3 answers · asked by emeka nnabuife 1 in Science & Mathematics ➔ Mathematics

lim (sinx+i)/(sinx-1)
= (sin0+i)/(sin0-1)
= i /-1
= -i

the end
.

2007-09-07 03:01:59 · answer #1 · answered by The Wolf 6 · 0⤊ 0⤋

This one's easy, since the function is actually continuous at 0:

[xâ0]lim (sin x + i)/(sin x - 1)
i/(-1)
-i

2007-09-07 02:17:03 · answer #2 · answered by Pascal 7 · 0⤊ 0⤋

sin 0=0
so the final answer is -i ..
juz substitute in sin x = sin 0.....

2007-09-07 03:08:07 · answer #3 · answered by wisten 2 · 0⤊ 0⤋