SREEKANTH
Last Activity: 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
so the answer is (n/m)^2