we can solve the problem using the series of log(1+x) and that of e^x..
let y=(1+x)^1/x
then, log y = 1/x log (1+x) (since log (1+x)=x - x^2/2 +x^3 /3...)
= 1/x[x - x^2/2+ x^3/3.......]
log y =1- x/2+ x^2/3................
therefore, y= e^1- x/2 + x^2/3 .....
y= e * e^ - x/2 + x^2/3....
y= e* [1+ (-x/2 +x^2/3) + (-x/2 +x^2/3)^2 /2] (since e^x=1+x+x^2/2!..)
y=e*[1-x/2 + 11x^2 /24]
Lt x tends to 0 [e(1-x/2 + 11x^2 /24) +ex - e] /x
Lt x tends to 0 [e - ex/2 + 11ex^2/24 + ex - e] /x
ans =e/2
