Numerator
Lim √(9x^6 - x)= (9x^6-x)^1/2
Denominator
Lim x^3 + 1
grade of numerator is (x^6)^ 1/2 = x^3 . IS THREE
And grade of numerator is also THREE
THEN IN THIS SPECIAL CASE, both have same grade
Then we have to DIVIDE numerator and denominator by x^3
Numerator/ x^3 = √(9x^ 6 - x)/ x^ 3 =
when x goes to infinite = √9
Denominator/x^3 = (x^3/x^3) + 1/x^3 = 1 + 1/x^3
when x goes to infinite = 1
Then NOW DIVIDE both results
= √ 9 / 1
= √9
So then ANSWER IS
lim √(9x^6 - x)/ x^3 + 1 = √9