How do you find the limit #(sqrt(x)-1)/(root3(x)-1)# as #x->1#?

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#

The #0/0# form invokes the use of L'Hôpital's rule

#(d(x^(1/2) - 1))/dx = 1/2x^(-1/2)#

#(d(x^(1/3) - 1))/dx = 1/3x^(-2/3)#

#lim_(xto1){1/2x^(-1/2)}/{1/3x^(-2/3)}#

#lim_(xto1) = 3/2x^(1/6) = 3/2#