Question 394624
We have {{{lim(x->9, (x^2 - 9x)/(2sqrt(x)-6))}}}. By direct substitution we get a fraction equal to 0/0 which is indeterminate. Using L'Hôpital's rule,


{{{lim(x->9, (x^2 - 9x)/(2sqrt(x) - 6)) = lim(x->9, (2x - 9)/(2(1/2sqrt(x)))) = lim(x->9, (sqrt(x))(2x-9)) = (sqrt(9))(2(9) - 9) = 27}}}