The limiting value of(cos x)1/sin xas~x→0 is
1
e
0
1/e
Put cos x = 1 + y as; x→0; y→0
Hence
limx→0 (cos x)1/sin x=limy→0[(1+y)1/y]y/sin xelimx→0cosx−1sin x=elimx→0(−tanx2)=e0=1
The limiting value of(cos x)1/sin xas x→0 is
limx→ 0loge(1+x)x=