Question

$\underset{x\to 0}{lim}\{\frac{{\log }_e(1+x)}{x^2}+\frac{x-1}{x}\}$ is equal to

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $1$
• D. none of these

Answer 1

Answer: (A)

$\frac{1}{2}$

${lim}_{x\to 0}\{\frac{{\log }_e(1+x)}{x^2}+\frac{x-1}{x}\}={lim}_{x\to 0}\left\{\frac{{\log }_e(1+x)+x^2-x}{x^2}\right\}={lim}_{x\to 0}\left\{\frac{(x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-...)+x^2-x}{x^2}\right\}=\frac{1}{2}$
Hence, option 'A' is correct.