Question

The value of $\underset{x\to 0}{lim}\frac{\log x}{x-1}$ using taylor series is?

• A. $1$
• B. $-1$
• C. $4$
• D. $-3$

Answer 1

The correct option is B $-1$

Taylor Series expansion of $\log x=(x-1)-\frac{1}{2}(x-1{)}^2+\frac{1}{3}(x-1{)}^3-...$

$\therefore \frac{\log x}{x-1}=1-\frac{1}{2}(x-1)+\frac{1}{3}(x-1{)}^2-..$

${lim}_{x\to 0}\frac{\log x}{x-1}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$

Thus ${lim}_{x\to 0}\frac{\log x}{x-1}=-1$