evaluate lim x tends to 0 (√1+x)-1/x

√(1+x)=(1+x)^1/2 = (1+x/2....)   (using binomial expansion)
==> lim_​x-->0 {(1+x/2)-1}/x
==> lim_​x-->0 {2+x-2}/2x      (Taking  LCM 2)
==> lim_​x-->0 {x}/2x
==> 1/2