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

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Answer 1 · 2017-04-15T01:28:45

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

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

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:

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

Answer 2 · 2017-04-15T07:37:32

$1$

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

so

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

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