Question

Find the value of limit x->0+ log (sin(x)) to the base sin(x/2)

Answer 1

Answer:

Let f(x)=xlog{sin(x)}

then, f(0.1)=0.1log{sin(0.1)} ; and we know for small values of x:sin(x)≈x

so, f(0.1)=0.1log{0.1}=0.1log{10−1}=−0.1

similarly, f(0.01)=0.01log{sin(0.01)}=0.01log{0.01}=0.01log{10−2}=−0.02

so we conclude as x→0 , f(x)→0

Therefore , limx→0+xlog{sin(x)} = 0