explain how as x approaches infinity of a square root?

Question

limit as x approaches infinity

f(x) = sq root of(9x^2-x) over (x^3+1) the answer is suppose to be 3 ... how? explain what to do with the square root, please?

Answer 1

f(x) = sqrt((9x^2-x)/(x^3+1)) = x/(x^3 + 1) * sqrt(9 - 1/x) = 1/( x^2 + 1/x) * sqrt(9 - 1/x)

As x --> oo, x^2 --> oo and 1/x --> 0, so that x^2 + 1/x --> oo and 1/( x^2 + 1/x) --> 0. And sqrt(9 - 1/x) -> sqrt(9 - 0) = sqrt(9) = 3. So, the answer is 0 * 3 = 0, not 3.

It would actulayy be 3 if the denominator was x +1 instead of x^3 + 1.

Answer 2

l'Hopital.