Let
#y=(e^x+x)^(1/x)#.
Generally, when we're dealing with a function in the exponent such as in this example, taking the logarithms and computing the limit of the logarithm as #x\to0# is much easier than trying to compute the limit directly.
Then we have that,
#ln(y) = 1/x * ln(e^x+x)#,
#ln(y) = 1/x ( ln(1+xe^(-x))+ln(e^x))#,
#ln(y) = 1/x (ln(1+xe^(-x))+x)#
#ln(y) = 1+ ln(1+xe^(-x))/x#
Power series also help in computing limits. Recall the series expansion for #ln(1+x)#, which holds for #abs(x)<1#.
#ln(1+x) = x - x^2/2 + x^3/3 - ...#.
Then, for #abs(xe^(-x)) < 1#,
#ln(1+xe^(-x)) = xe^(-x) - (xe^(-x))^2/2+(xe^(-x))^3/3-...#
This expansion holds for #x=0#, as #abs(0*e^(-0))<1#.
Then,
#lim_{x\to0} ln(y) = lim_{x\to0} 1 + (xe^(-x) - (xe^(-x))^2/2+(xe^(-x))^3/3-...)/x#,
#lim_{x\to\0} ln(y) = lim_{x\to0} 1 + e^(-x) - (xe^(-2x))/2+(x^2e^(-3x))/3+...#.
This right hand side can now be easily evaluated.
#lim_{x\to0} ln(y) = 1+e^(-0)#,
#lim_{x\to0} ln(y) = 2#.
So, as #x# goes to 0, the logarithm of #y# goes to #2#. By taking an inverse logarithm,
#lim_{x\to0} y = e^2#.
