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

#(e^x+x)^(1/x)=e(1+x/e^x)^(1/x)# now making #y=x/e^x# we have
#e(1+y)^(e^(-x)/y)=e((1+y)^(1/y))^(e^(-x))# here
#{(x->oo),(y(x)->0):}# then

#lim_(x->oo)(e^x+x)^(1/x)=e lim_(x->oo)(lim_(y->0)(1+y)^(1/y))^(e^(-x))=e cdot e^0=e#

#L=lim_(xrarroo)(e^x+x)^(1/x)#

A helpful tool in finding limits with exponential functions like this is to use the equivalence #a=e^lna#. Thus

#L=lim_(xrarroo)e^(ln(e^x+x)^(1/x))#

This can be rewritten using the log rule #log(a^b)=bloga# as

#L=lim_(xrarroo)e^(1/xln(e^x+x))#

We can move the limit inside of the exponential function:

#L=e^(lim_(xrarroo)(ln(e^x+x)/x)#

Notice that #lim_(xrarroo)ln(e^x+x)/x# is in the indeterminate form #oo/oo#, so L'Hopital's rule applies. We can take the derivatives of the numerator and denominator:

#L=e^(lim_(xrarroo)((d/dxln(e^x+x))/(d/dxx))#

#L=e^(lim_(xrarroo)((e^x+1)/(e^x+x))/1#

#L=e^(lim_(xrarroo)(e^x+1)/(e^x+x)#

You may immediately recognize that this limit is #1#, since the #e^x# terms in the numerator and denominator will overpower the other terms, so as #x# approaches infinity, #(e^x+1)/(e^x+x)approxe^x/e^x=1#, but in case this isn't enough for you we can do L'Hopital's again since we are in the #oo/oo# form.

#L=e^(lim_(xrarroo)(e^x/(e^x+1))#

And L'Hopital's once more, as we're still in #oo/oo#:

#L=e^(lim_(xrarroo)(e^x/e^x)#

#L=e^(lim_(xrarroo)1)#

#L=e^1#

#L=e#