What is the limit of #(x/(x+1))^x # as x approaches infinity?

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Answer 1 · 2016-03-21T20:55:02

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

First, we will use the following:

#e^ln(x) = x#
Because #e^x# is continuous on #(-oo,oo)#, we have #lim_(x->oo)e^f(x) = e^(lim_(x->oo)f(x))#

With these:

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

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

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

Next, we will use L'Hopital's rule:

#lim_(x->oo)xln(x/(x+1)) = lim_(x->oo)ln(x/(x+1))/(1/x)#

#=lim_(x->oo)(d/dxln(x/(x+1)))/(d/dx1/x)#

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

#lim_(x->oo)-x/(x+1)#

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

#=-1#

Putting them together, we get our final result.

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