#lim_(xrarroo)(sqrt((x^2-1)/(x^2+1)))^((2x+3)/(x-2))#
Use exponential function and the natural logarithm:
#lim_(xrarroo)e^ln((sqrt((x^2-1)/(x^2+1)))^((2x+3)/(x-2)))#
Use the property #ln(a^c) = (c)ln(a)#
#lim_(xrarroo)e^(((2x+3)/(x-2))ln((sqrt((x^2-1)/(x^2+1)))))#
Do that again for the square root:
#lim_(xrarroo)e^(1/2((2x+3)/(x-2))ln((x^2-1)/(x^2+1)))#
Multiply the #1/2# into the fraction:
#lim_(xrarroo)e^(((2x+3)/(2x-4))ln((x^2-1)/(x^2+1)))#
Use of L'Hôpital's rule shows that both #((2x+3)/(2x-4))# and #(x^2-1)/(x^2+1) to 1" as " x to oo#
#((d(2x+3))/dx)/((d(2x-4))/dx) = 2/2 = 1#
#((d(x^2-1))/dx)/((d(x^2-1))/dx) = (2x)/(2x) = 1#
#e^(1ln(1)) = 1#
Therefore, the limit of the original expression is 1:
#lim_(xrarroo)(sqrt((x^2-1)/(x^2+1)))^((2x+3)/(x-2)) = 1#
