Limit of x^1/x as x Approaches Infinity: Simplified Using ln and e

[golriz]

Hello
please help me and compute this limit:
llim x^1/x when x approaches infinite.
I think we can rewrite this so: e^ln (x^1/x).

 

[oma12]

ok so it simplifies to e^((lnx)/x) and the limit of (lnx)/x is 0, so? e^0 = 1

 

[golriz]

thanks

 

[golriz]

thanks
it was very easy!