limx→0 xloge(sinx) is equal to
-1
loge1
1
none of these
We have,
limx→0 xlogesinx [0×∞ form ]
=limx→0 logesinx 1/x ∞∞ form
=limx→0 cotx−1/x2 [Using L' Hospital's rule]
=−limx→0 x2tanx=−limx→0 xtanx×x=−(1×0)=0=loge1