Given: #lim_(x to 1) 1/(sqrt (x-3) + sqrt 3)#
#lim_(x to 1) 1/(sqrt (x-3) + sqrt 3) = 1/(sqrt (1-3) + sqrt 3)#
#lim_(x to 1) 1/(sqrt (x-3) + sqrt 3) = 1/(sqrt (-2) + sqrt 3)#
#lim_(x to 1) 1/(sqrt (x-3) + sqrt 3) = 1/(sqrt3+ sqrt2i)#
Multiply by 1 in the form of #(sqrt3- sqrt2i)/(sqrt3- sqrt2i)#:
#lim_(x to 1) 1/(sqrt (x-3) + sqrt 3) = 1/(sqrt3+ sqrt2i)(sqrt3- sqrt2i)/(sqrt3- sqrt2i)#
We do this because we know that multiplication by the conjugate converts the denominator into a real number.
#lim_(x to 1) 1/(sqrt (x-3) + sqrt 3) = (sqrt3- sqrt2i)/(3- 2)#
#lim_(x to 1) 1/(sqrt (x-3) + sqrt 3) = (sqrt3- sqrt2i)/1#
#lim_(x to 1) 1/(sqrt (x-3) + sqrt 3) = sqrt3- sqrt2i#