limx→11−xln(x)=\lim_{x \to 1} \frac{1-x}{ln(x)}=limx→1ln(x)1−x= A0B1/2C1D-1Answer: D. -1 Read Explanation: limx→11−xln(x)=\lim_{x \to 1} \frac{1-x}{ln(x)}=limx→1ln(x)1−x=applying L Hospitals rulelimx→11−xln(x)=limx→1−11x=limx→1(−x)=−1\lim_{x \to 1} \frac{1-x}{ln(x)}= \lim_{x \to 1} \frac{-1}{\frac{1}{x}} = lim_{x \to 1} (-x) = -1limx→1ln(x)1−x=limx→1x1−1=limx→1(−x)=−1 Read more in App
