$f\left ( x \right ) = (1 - \frac{1}{n})^{2n}$
f(x) is of the form $1^{\infty }$
Taking y = f(x),
$\ln y = 2n*\lim_{n \to \infty }\ln (1-\frac{1}{n})$
$\Rightarrow $$\ln y = 2* \lim_{n \to \infty }\frac{\ln ({1 - \frac{1}{n})}}{1/n}$ ($\frac{0}{0}$ form)
Applying L'Hospitals rule,
$\ln y = 2* \lim_{n \to \infty }(\frac{\frac{1}{1 - \frac{1}{n}}*(-\frac{1}{n^{2}})}{-1/n^{2}})$
$\Rightarrow \ln y = -2*\lim_{n \to \infty }(\frac{1}{1-\frac{1}{n}})$
$\Rightarrow \ln y = -2$
$\Rightarrow y = e^{-2}$
Hence option B.