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

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Answer 1 · 2017-03-05T16:18:43

$1/e^2$

Solving for $y$

$(x - 1)/(x + 1) = 1 + 1/y$

we obtain $y=-1/2(x+1)$ and $x = - (2 y+1)$

also $lim_(x->oo) equiv lim_(y->-oo)$ then

$lim_(x->oo)((x - 1)/(x + 1) )^x equiv lim_(y->-oo)(1+1/y)^(-(2y+1))$

or

$lim_(x->oo)((x - 1)/(x + 1) )^x equiv lim_(z->oo)(1+1/(-y))^(2y-1) =$

$= lim_(z->oo)(1+1/(-y))^-1 lim_(y->oo)(1+1/(-y))^((-2)(-y)) = 1 cdot e^(-2)$

so

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

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