#lim_(x->oo)(sqrt((x^2-1)/(x^2+1)))^((2x+3)/(x-2))=# ?

2 Answers
Apr 15, 2017

#lim_(xrarroo)(sqrt((x^2-1)/(x^2+1)))^((2x+3)/(x-2))=1#

Explanation:

#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#

Apr 15, 2017

#1#

Explanation:

#((x^2-1)/(x^2+1))^(1/2(2x+3)/(x-2))=((1-1/x^2)/(1+1/x^2))^(1/2((2+3/x)/(1-2/x)))=((1-1/x^2)/(1+1/x^2))^((1+3/(2x))/(1-2/x))#

so

#lim_(x->oo)(sqrt((x^2-1)/(x^2+1)))^((2x+3)/(x-2))=lim_(x->oo)((1-1/x^2)/(1+1/x^2))^((1+3/(2x))/(1-2/x))=1^1=1#