# limit x tends to 0 then what is the value of 1-cosmx/1-cosnx=

PASUPULETI GURU MAHESH
126 Points
8 years ago
limit x tends to 0   1-cosmx/1-cosnx= now rationalise nx))/(1-cos2nx)(1+cosnx)= ((1-cosmx)x1+cosnx/)x(1+cosnx(1-cosmx/1-cos
=(1+cosnx-cosmx.cosnx-cosmx)/(1-cos2nx)
=((1+cosnx).(1-cosmx))/(1-cos2nx)
=(1+cosnx)/(1-cos2nx)×(1-cosmx)
then solve that you get zerothere fore limit x tends to 0 then what is the value of 1-cosmx/1-cosnx=0
DEEPTHI JANGA
51 Points
8 years ago
(1-cosmx/1-cosnx)x(1+cosnx/1+cosnx)= ((1-cosmx)x(1+cosnx))/(1-cos2nx)1-cosmx/1-cosnx= now rationalise tends to 0   limit x
=(1+cosnx-cosmx.cosnx-cosmx)/(1-cos2nx)
=((1+cosnx).(1-cosmx))/(1-cos2nx)
=(1+cosnx)/(1-cos2nx)×(1-cosmx)
then solve that you get zerothere fore limit x tends to 0 then what is the value of 1-cosmx/1-cosnx=0 sorry but the options are a.mn  b.m/n   c.m2/n2    d.m-n
SREEKANTH
85 Points
8 years ago
limit x tends to zero (1-cosmx)/(1-cosnx) by dividing the euation with mx and multiply with nx then you will get =limit x tends to zero (1-cosmx)/(1-cos nx)*(nx)/(mx) on re arranging we will get
=((mx)(1-cosmx)/(mx)/(nx)(1-cos nx)/(nx)  but we know that 1-cos x=2sin^2 x/2
=(nx/mx)(2(2sin^2 (mx)/2)/(mx)/2)/2(2sin^2 (nx)/2)/(nx)/2)
=(n/m)(16/16)(n/m) limit x tends to zero ((sin (mx)/2)/(mx)/2))sin (mx)/2)/(mx)/2))/sin^2 (nx)/2)/(nx)/2)sin^2 (nx)/2)/(nx)/2)
=(n/m)^2
Ajay
209 Points
8 years ago
Since this of form 0/0 use L Hospital rule. Differentiating numerator and denominator it becomes
(m sin  mx)/ (n sin nx)
This is again of form 0/0 , hence differentiating again
(m2 cos mx)/ (n2 cos nx)
as x tends to 0,  both cos mx and con nx tends to 1 hence m2/n should be correct answer