Question

$\underset{x\to 0}{lim}\left\{\right(1+x{)}^{\frac{2}{x}}\}$ (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 $e^2-7$

As $\left\{\right(1+x{)}^{\frac{2}{x}}\}=(1+x{)}^{\frac{2}{x}}-[(1+x{)}^{\frac{2}{x}}]$
And $\underset{x\to 0}{lim}(1+x{)}^{\frac{2}{x}}=exp\underset{x\to 0}{lim}\left(\frac{2}{x}\log (1+x)\right)=exp\underset{x\to 0}{lim}\left(\frac{2\frac{1}{1+x}}{1}\right)=e^2$

$Since$ $[\underset{x\to \infty }{lim}{f\left(x\right)}^{g\left(x\right)}=e^{(\underset{x\to \infty }{lim}g\left(x\right).lnf\left(x\right))}]$
$\therefore \underset{x\to 0}{lim}\left\{\right(1+x{)}^{\frac{2}{x}}\}=e^2-\left[e^2\right]=e^2-7$