Quantative aptitude problems

Let √x be t now the limit function can be written as (t³ – 27)/(t² – 9) In the above expression,  t-3 is the common term cancelling it we get  (t² + 3t + 9)/(t+3) when x tends to 9 t tends to 3 so limit is (9 + 9 + 9)/(3+3) = 27/6 =4.5

thanks..

lim    ( x^n-a^n)/x-a=na^(n-1)

x -> a

lim (x^3/2-27)/x-9 = 3/2*9^(3/2-1) = 4.5

x->9