Question

Evaluate: $\underset{x\to 0}{lim}[\frac{1}{x}-\frac{\log (1+x)}{x^2}]$

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $1$
• D. $-1$

Answer 1

The correct option is A $\frac{1}{2}$

We have $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-......$ for $-1<x\le 1$

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

or, $[\frac{1}{x}-\frac{\left(\log \right(1+x\left)\right)}{x^2}]=\frac{1}{2}-\frac{x}{3}+.....$

or, $\underset{x\to 0}{lim}[\frac{1}{x}-\frac{\left(\log \right(1+x\left)\right)}{x}]=\underset{x\to 0}{lim}[\frac{1}{2}-\frac{x}{3}+.....]$

or, $\underset{x\to 0}{lim}[\frac{1}{x}-\frac{\left(\log \right(1+x\left)\right)}{x^2}]=\frac{1}{2}$