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

Question

4241 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-19T03:23:24

Rationalize and reduce.

#(sqrtx-1)(sqrtx+1) = x-1# #" "# (use #a^2-b^2=(a-b)(a+b)#)

and

#(root3x-1)(rootrx^2+root3x+1) = (x-1)# (use #a^3-b^3 = (a-b)(a^2+ab+b^2)#)

So,

#((root3 x) -1)/((sqrt x) -1) = ((x-1)(sqrtx+1))/((x-1)(rootrx^2+root3x+1))#

# = (sqrtx+1)/(rootrx^2+root3x+1)#

Taking the limit as #xrarr1# yields #2/3#.