Question

$\underset{x\to 0}{lim}[\frac{1}{x}-\frac{lo{g}_e(1+x)}{x^2}]=$

• A.
• B.
• C. 1
• D. -1

Answer 1

The correct option is A

We have $\underset{x\to 0}{lim}[\frac{1}{x}-\frac{lo{g}_e(1+x)}{x^2}]$

At x=0,$\underset{x\to 0}{lim}\frac{log(1+x)}{x}=1$

At x=0 above given expression take $\infty -\infty$ form.

Expansion of $lo{g}_e(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}........$

$\underset{x\to 0}{lim}[\frac{1}{x}-\frac{x-\frac{x^2}{2}+\frac{x^3}{3}-........}{x^2}]$

$\underset{x\to 0}{lim}[\frac{1}{x}-(\frac{1}{x}-\frac{1}{2}+\frac{x}{3}-........)]$

$\underset{x\to 0}{lim}[\frac{1}{x}-\frac{1}{x}+\frac{1}{2}-\frac{x}{3}+........]$

$=\frac{1}{2}-0$

$=\frac{1}{2}$