Question

$\underset{x\to 0}{lim}\left\{{(1+x)}^{\frac{2}{x}}\right\}$, where $\{.\}$ denotes the fractional part of $x$, is equal to

• A. $e^2-7$
• B. $e^2-8$
• C. $e^2-6$
• D. None of these

Answer 1

The correct option is A $e^2-7$

${lim}_{x\to 0}\left\{{(1+x)}^{\frac{2}{x}}\right\}$
${lim}_{x\to 0}({(1+x)}^{\frac{2}{x}}-\left[{(1+x)}^{\frac{2}{x}}\right])(\because x-\left[x\right]=\left\{x\right\})[-]$ denotes greatest integer
$={lim}_{x\to 0}{(1+x)}^{\frac{2}{x}}-{lim}_{x\to 0}\left[{(1+x)}^{\frac{2}{x}}\right]$
$=e^{{lim}_{x\to 0}(1+x-1)\frac{2}{x}}-\left[e^{\${lim}_{x\to 0}(1+x-1)\frac{2}{x}}\right]$
$=e^2-\left[e^2\right](\because {lim}_{x\to 0}{(1+x)}^{\frac{2}{x}}=1^{\infty }form;\therefore {lim}_{x\to 0}{[1+f\left(x\right)]}^{\frac{c}{f\left(x\right)}}=e^{{lim}_{x\to 0}c})$
Value of $e=2.718$
$=e^2-\left[{\left(2.718\right)}^2\right]$
$=e^2-\left[7.387\right]$
$=e^2-7$