Finding a limit with a Square Root

Question

$$\lim_{x\to \infty} \frac{\sqrt{9x^6-x}}{x^3+7}$$

I thought it would simply be $1/3$, not sure where I went wrong.

Answer 1

$$\lim_{x\to \infty}\frac{\sqrt{9x^6 - x}}{x^3 + 7}\quad = \quad\lim_{x\to \infty} \frac{\sqrt{9 - \frac{1}{x^5}}}{1 + \frac 7{x^3}} = \frac {\sqrt{9 - 0}}{1 + 0} = \sqrt 9 = 3$$

I divided the numerator and denominator by $x^3$.

Answer 2

Since $$x=_\infty o(x^6)\quad;\quad 7=_\infty o(x^3)$$ then we can write $$\require{cancel}\lim_{x\to-\infty}\frac{\sqrt{9x^6-x}}{x^3+7}=\lim_{x\to-\infty}\frac{\sqrt{9x^6\cancel{-x}}}{x^3\cancel{+7}}=\lim_{x\to-\infty}\frac{\overbrace{-3x^3}^{\ge0}}{x^3}=-3$$