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

Answer 1

Evelyn Agard

#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))#

#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)#

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

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Answer 2

Ezekiel Alexander

The limit of ((x - 1)/(x + 1))^x as x approaches infinity is equal to e^(-2).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

#lim_(x-&gt;oo)((x - 1)/(x + 1) )^x =# ?

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