limx→0x loge x will be equal to
We can observe that
As x→0 ,logex→−∞
So if we write the above expression as logex1x then,
as x→0, 1x→∞;logex1xreduces to (∞∞) form.
So we can use L’Hospital Rule. We are doing all this because direct substitution of x as zero is not giving us the limit.
So L’Hospital Rule states that limx→af(x)g(x)=limx→af′(x)g′(x)
So for logex1x taking derivative of numerator and denominator we get 1x−1x2
So limx→0logex1x=limx→01x−1x2=limx→0(−x)=0 which is the correct answer.