Limit to infinity

**PhizKid** · October 29th 2012, 03:58 AM

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

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?

**a tutor** · October 29th 2012, 04:26 AM

Take your denominator into the square root $\sqrt{\frac{9x^6-x}{(x^3+1)^2}}$ .

You may see that the numerator and denominator both have 6 as their highest power of x.

Divide top and bottom by x^6.

Thread: Limit to infinity

LinkBack

Thread Tools

Display

Limit to infinity

Re: Limit to infinity

Similar Math Help Forum Discussions

If the derivative is bigger than positive constant, is the limit at infinity infinity

Definition of Limit proof: x -> infinity of sqrt(x) - x = -infinity

Limit at infinity equaling infinity?

limit to infinity !!

Limit of ln(x+1) as x->infinity

Search Tags