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