It would help if someone could explain why.
Thank you.
2006-11-26 07:38:09 · 6 answers · asked by Bender[OO] 3 in Science & Mathematics ➔ Mathematics
Edit: Ah yes the question was a limit problem dealing with L'hopitals rule and it was the limit as x approached 1 from the right, i forgot completely about the little positive sign. Alright now i see why it would be negative infinity when it approaching 1 from the right. Thanks for your quick answers!
2006-11-26 07:46:08 · update #1
Depends
lim θ→π/2- (ie from below) tanθ = +∞
lim θ→π/2+ (ie from above) tanθ = -∞
This is because in the 1st quadrant tanθ >0 and in the second quadrant tanθ <0
2006-11-26 07:40:31 · answer #1 · answered by Wal C 6 · 0⤊ 0⤋
tanx= sinx / cos x
sin(pi/2)=1
cos(pi/2)=0
tan(pi/2)=sin(pi/2) / cos(pi/2).
Numerator is 1. Tan (pi/2) gets larger and larger as the angle gets closer to pi/2.
So tan(pi/2) approaches positive infinity, not negative infinity.
2006-11-26 15:44:43 · answer #2 · answered by fcas80 7 · 0⤊ 0⤋
Tan pi/2 = sin(pi/2)/cos(pi/2) at pi/2 the denominator is zero...
As you approach pi/2 from the low side tan pi/2 -> + infinity
as you approach pi/2 from the high side tan pi/2 -> - infinity
It is undefined at pi/2.
If you have a graphing calculator, I use a TI-83, you can graph it.
2006-11-26 15:41:25 · answer #3 · answered by modulo_function 7 · 0⤊ 0⤋
- infinity= -1(infinity) =infinity
So if you say tan Pi/2= - infinity that is no problem.
2006-11-26 15:57:57 · answer #4 · answered by aminnyus 2 · 0⤊ 0⤋
no +ve infinity
2006-11-26 15:41:15 · answer #5 · answered by raj 7 · 0⤊ 0⤋
why are you eating Tan-Ï? I like mine to be lightly brown(like my men, ha, AIRPLANE REFERENCEISH THINGY), anyway um
there;s no need to explain
I just eat my pie whenever I
feel like it
2006-11-26 15:41:14 · answer #6 · answered by Anonymous · 1⤊ 1⤋