Given #h(x)=(e^x+x)^(1/x)#

First, we'll check how the base comes out if we apply the limit #x\to0#. It seems that if we do so, then #e^x+x=e^0+0=1#

But, check the power, it seems that then the equation would become #1^\infty#. We could leave it at that but since this is limits, we need to be careful here.

So, to solve this we have an equation, that is if #h(x)=g(x)^f(x)# and as #x\toa# such that #g(x)\to0# and #f(x)\toinfty#, then we have a solution which is

#lim_{x\toa}h(x)=e^(h(x)*(g(x)-1)#

So, substituting #g(x)=e^x+x# and #f(x)=1/x#, we get

#lim_{x\to0}h(x)=e^(1/x(e^x+x-1)#

Let's now just take care of the powers, so we get

#lim_{x\to0}(e^x+x-1)/x=lim_{h\to0}(e^x-1)/x+x/x#

We know about the #(e^x-1)/x# at #x# tends to #0# limit equaling #1#, so we get it in the end as #1+1=2#

So, that is why there's an #e# in the answer.