lim x-> 0 sin2x/tanx
i got 2 for an answer but i have this nagging feeling that i did it wrong.
2007-03-03 03:59:46 · 6 answers · asked by googoogoo 2 in Science & Mathematics ➔ Mathematics
Answer =2.
sin 2x = 2 sinx cosx.
So your quotient = 2 sinx cosx / (sinx/cosx) =
2 cosx cosx --> 2 * 1 * 1 = 2
2007-03-03 04:06:58 · answer #1 · answered by Anonymous · 0⤊ 0⤋
lim x->0 sin2x/tanx = limx->0 (sin2x)'/(tanx)' where the ' means the derivative of of each expression in brackets. Thus
linx->0 (2cos2x)/sec^2x = linx->0 (2cos2x)cos^2x = 2x1x1 = 2
2007-03-03 12:08:56 · answer #2 · answered by physicist 4 · 0⤊ 0⤋
lim x->0 sin2x/tanx=lim x->0 (2sinxcosx)/tanx
=lim x->0 (2sinxcosx)/(sinx/cosx)
=lim x->0 2cos^2x
=2cos^20
=2
2007-03-03 12:07:01 · answer #3 · answered by bksrikanth 1 · 0⤊ 0⤋
sin2x/tanx = (sin2x /2x ) *2x /(tanx/x)*x
As the parenthesys have limit 1(each) and the x simplifies the limit is 2
2007-03-03 12:08:05 · answer #4 · answered by santmann2002 7 · 0⤊ 0⤋
sin(2x)=2* sin(x)*cos(x)
tan(x)=sin(x)/cos(x)
So: sin2x/tanx = 2[sinx*cosx]/[sinx/cosx]
=2(cosx)^2
SInce lim as x->0 of cosx is 1,
then lim [2(cosx)^2] = 2.
Done.
2007-03-03 12:25:42 · answer #5 · answered by Jerry P 6 · 0⤊ 0⤋
its right!
2007-03-03 12:02:25 · answer #6 · answered by FedUp 3 · 0⤊ 0⤋