what will be the limit of lt x-0 (sinx^o/x)
2006-09-06 05:30:40 · 4 answers · asked by sweepie 2 in Science & Mathematics ➔ Mathematics
limit sin x^o / x
x=>0
First change degrees into radians by using the formula
180^o = pi radians
x^o =(pi/180)x
limit sin {(pi/180)x } / x
x=>0
limit [sin {(pi/180)x }]{180/pi} / x{(pi/180)
x=>0
= pi/180 when x=>0
Hence (pi/180)
2006-09-06 06:00:14 · answer #1 · answered by Amar Soni 7 · 0⤊ 1⤋
I think x^o means "x degrees", and "lt x-0" means "limit as x approaches zero". So I can rewrite it as
lim (x ==> 0) (sin x)/x ==> 0/0 (because sin 0 = 0)
So take the derivative of the top & bottom (L'Hopital):
lim (x ==> 0) (cos x)/1 = 1 (answer) (because cos 0 = 1)
2006-09-06 14:28:27 · answer #2 · answered by bpiguy 7 · 0⤊ 0⤋
That's a good question. But before you expect anyone to answer, you may want to write your formula in a way that someone would be able to read without misunderstandings.
2006-09-06 12:34:49 · answer #3 · answered by Anonymous · 1⤊ 0⤋
Only you and your teacher will know
2006-09-06 12:33:35 · answer #4 · answered by Anonymous · 0⤊ 0⤋