Form :- (→1)$^{→\infty}$ => indeterminate form.
$\lim_{n→\infty}$ (1 – 1/n)$^{2n}$
= $\lim_{n→\infty}$ e$^{(ln(1 – 1/n)^{2n})}$
= $\lim_{n→\infty}$ e$^{(2n)ln(1 – 1/n)}$ (let’s find $\lim_{n→\infty}$ n ln(1 – 1/n))
$\lim_{n→\infty}$ n ln(1 – 1/n) (→$\infty$ x →0 => indeterminate form)
= $\lim_{n→\infty}$ [ln(1 – 1/n)] / (1/n) (→$\infty$/→$\infty$ => indeterminate form. Apply L’hopital’s rule)
= $\lim_{n→\infty}$ [1/n$^{2}$] / (1 – 1/n)(-1/n$^{2}$)
= $\lim_{n→\infty}$ (-1)/(1 – 1/n)
= -1
So, $\lim_{n→\infty}$ e$^{(2n)ln(1 – 1/n)}$ = $\lim_{n→\infty}$ e$^{(2)[nln(1 – 1/n)]}$ = $\lim_{n→\infty}$ e$^{(2)(-1)}$ = e$^{-2}$
Another approach :- Use formula : $\lim_{n→a}$ f(n)$^{g(n)}$ where when n→a, then f(n)→1 and g(n)→$\infty$ ((→1)$^{→\infty}$ => indeterminate form)
$\lim_{n→a}$ f(n)$^{g(n)}$ = $\lim_{n→a}$ e$^{g(n)(f(n)–1)}$
$\lim_{n→\infty}$ (1 – 1/n)$^{2n}$ = $\lim_{n→\infty}$ e$^{2n(1 – (1/n) –1)}$ = $\lim_{n→\infty}$ e$^{2n(-1/n)}$ = $\lim_{n→\infty}$ e$^{-2}$ = e$^{-2}$