i will solve by two methods
method 1:
$y =\lim\limits_{n \to \infty}\left(1-\frac{1}{n}\right)^{2n}$
Taking log
$\log y =\lim\limits_{n \to \infty} 2n \log \left(1-\frac{1}{n}\right)$
$ =\lim\limits_{n \to \infty} \dfrac{\log \left(1-\frac{1}{n}\right)}{\left(\frac{1}{2n}\right)}$---------------(converted this so as to have form $\left(\frac{0}{0}\right)$)
apply L' hospital rule
$\log y=\lim\limits_{n \to \infty} \dfrac{\left(\dfrac{1}{1-\frac{1}{n}}\right).\frac{1}{n^{2}}}{\left(\dfrac{-1}{2n^{2}}\right)}$
$\log y={-2}$
$y=e^{-2}.$
method 2:
it takes $1$ to power infinity form
$\lim\limits_{x \to \infty} f(x)^{g(x)}$
$=e^{\lim\limits_{x \to \infty} (f(x)-1)g(x)}$
where, $(f(x)-1)*g(x)=\frac{-1}{n}*{2n}={-2}.$
ie -2 constant.
so we get final ans is $= e^{-2}.$
u can refer this link for second method
http://www.vitutor.com/calculus/limits/one_infinity.html
Correct Answer: $B$