What we can do here is fairly unintuitive. Recall that we can use the difference of cubes identity, or #a^3-b^3=(a-b)(a^2+ab+b^2)# to show that #x-1=(root3x-1)(root3(x^2)+root3x+1)#.
So, we can multiply the function by what could be considered its "cubic conjugate:"
#lim_(xrarr1)(sqrtx-1)/(root3x-1)=lim_(xrarr1)(sqrtx-1)/(root3x-1)*(root3(x^2)+root3x+1)/(root3(x^2)+root3x+1)#
#color(white)(lim_(xrarr1)(sqrtx-1)/(root3x-1))=lim_(xrarr1)((sqrtx-1)(root3(x^2)+root3x+1))/(x-1)#
We can also multiply by the conjugate of the term with the square root:
#color(white)(lim_(xrarr1)(sqrtx-1)/(root3x-1))=lim_(xrarr1)((sqrtx-1)(root3(x^2)+root3x+1))/(x-1)*(sqrtx+1)/(sqrtx+1)#
#color(white)(lim_(xrarr1)(sqrtx-1)/(root3x-1))=lim_(xrarr1)((sqrtx-1)(sqrtx+1)(root3(x^2)+root3x+1))/((x-1)(sqrtx+1))#
#color(white)(lim_(xrarr1)(sqrtx-1)/(root3x-1))=lim_(xrarr1)((x-1)(root3(x^2)+root3x+1))/((x-1)(sqrtx+1))#
#color(white)(lim_(xrarr1)(sqrtx-1)/(root3x-1))=lim_(xrarr1)(root3(x^2)+root3x+1)/(sqrtx+1)#
#color(white)(lim_(xrarr1)(sqrtx-1)/(root3x-1))=(1+1+1)/(1+1)#
#color(white)(lim_(xrarr1)(sqrtx-1)/(root3x-1))=3/2#
