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

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Answer 1 · 2016-11-29T02:52:31

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

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

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

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

The above step is valid due to the continuity of #f(x)=e^x#

Evaluating the limit in the exponent, we have

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

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

The above step follows from apply L'Hopital's rule to a #oo/oo# indeterminate form

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

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

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

#=1/(1+1/oo)#

#=1/(1+0)#

#=1#

Substituting this back into the exponent, we get our final result:

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

#=e^1#

#=e#