ISI-SAMPLE-2014

Question

Answer 1

Answer is (D)1

Answer 2

$\lim_{n\rightarrow \infty }{(1 - 1/n^2)}^n$

This is of $1^\infty$,   

for this we need to use  $\lim_{n\rightarrow \infty}{f(x)}^{g(x)} = e^{ \lim_{n\rightarrow \infty}(f(x)-1)*g(x)}$

$f(x) = (1 - 1/n^2)$

$g(x) = n$

$= e^{ \lim_{n\rightarrow \infty}(1 - 1/n^2-1)*n}$

$= e^{ \lim_{n\rightarrow \infty} - 1/n}$

$= e^{ - 1/\infty }$

$= e^{ 0 }$

$= 1$