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

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Answer 1 · 2018-02-18T17:42:49

$e^-2$

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

now making $y =- (x+1)/2$ we have

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

but

$(1+1/y)^(-(2y+1))=1/(1+1/y)(1+1/y)^(-2y)$ and then

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

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