Making #y = 4n+5# we have
#((y-5)/y)^((y-5)/2) = (1-5/y)^(y/2) (1-5/y)^(-5/2)#
now making #xi = -y/5# and substituting
# (1-5/y)^(y/2) (1-5/y)^(-5/2) = (1+1/xi)^(-5/2 xi)(1+1/xi)^(-5/2) =#
# = ( (1+1/xi)^xi)^(-5/2)(1+1/xi)^(-5/2)#
Now #n->oo rArr y->oo rArr xi -> -oo# and
#lim_(n->oo)((4 n)/(4 n + 5))^(2 n) = lim_(xi->-oo)( (1+1/xi)^xi)^(-5/2)(1+1/xi)^(-5/2) = #
#(lim_(xi->-oo) (1+1/xi)^xi)^(-5/2)lim_(xi->oo)(1+1/xi)^(-5/2) = e^(-5/2)#
NOTE
We were using the fact
#lim_(xi->-oo) (1+1/xi)^xi = lim_(xi->oo) (1+1/xi)^xi = e#