Solving \lim_{x \to \infty} \frac{\sqrt{9x^6 - x}}{x^3 + 1}

[PhizKid]

Homework Statement

$$\lim_{x \to \infty} \frac{\sqrt{9x^6 - x}}{x^3 + 1}$$

Homework Equations

The Attempt at a Solution

$$\frac{\sqrt{9x^6 - x}}{x^3 + 1} \cdot \frac{\frac{1}{x^3}}{\frac{1}{x^3}} = \\ \frac{\frac{\sqrt{9x^6 - x}}{x^3}}{1 + \frac{1}{x^3}} =\\ \frac{(1 + \frac{1}{x^3})(\sqrt{9x^6 - x})}{x^3} = \\ \frac{(1 + \frac{1}{x^3})(\sqrt{9x^6 - x})}{x^3} \cdot \frac{\frac{1}{x^3}}{\frac{1}{x^3}} = \\ \frac{(1 + \frac{1}{x^3})(\sqrt{9x^6 - x})}{x^3}$$

And it just repeats over and over again and I can't find anything to divide by without destroying the work I've already done. What am I supposed to do in a loop and there's nothing to divide by?

 

[clamtrox]

You have to keep better track of the order of calculations.. Remember that a/(b/c) ≠ (a/b)/c.

 

[SammyS]

Homework Statement

$$\lim_{x \to \infty} \frac{\sqrt{9x^6 - x}}{x^3 + 1}$$

Homework Equations

The Attempt at a Solution

$$\frac{\sqrt{9x^6 - x}}{x^3 + 1} \cdot \frac{\frac{1}{x^3}}{\frac{1}{x^3}} = \\ \frac{\frac{\sqrt{9x^6 - x}}{x^3}}{1 + \frac{1}{x^3}} =\\$$

The next line is not equivalent to the above.

Then use the fact that $x^3=\sqrt{x^6}\ .$

$$\frac{(1 + \frac{1}{x^3})(\sqrt{9x^6 - x})}{x^3} = \\ \frac{(1 + \frac{1}{x^3})(\sqrt{9x^6 - x})}{x^3} \cdot \frac{\frac{1}{x^3}}{\frac{1}{x^3}} = \\ \frac{(1 + \frac{1}{x^3})(\sqrt{9x^6 - x})}{x^3}$$

And it just repeats over and over again and I can't find anything to divide by without destroying the work I've already done. What am I supposed to do in a loop and there's nothing to divide by?

 

[mfb]

Here is a hint: Try to combine the 1/x^3 with your square root.