How do you find the limit of #(1+2/x)^x# as x approaches #oo#?

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Answer 1 · 2016-04-12T01:59:11

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

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

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

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

(note that the last step is valid because #e^x# is continuous)

#lim_(x->oo)xln(1+2/x) = lim_(x->0)ln(1+2x)/x#

#=lim_(x->0)(d/dxln(1+2x))/(d/dxx)#

#=lim_(x->0)2/(1+2x)#

#=2#

Substituting this into #"( * )"# we get our result:

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