the problem is

the lim as x approaches inifinity of

√(9x^6 - 1)/(x^3 - 1)

the 1's are constants being subtracted, not being subtracted from the exponents themselves.

I tried mulitplying both by 9x^6-1 but it just wasn't a good time and I couldn't see it ever working out.

I tired multiplying by 1/x but that wasn't really a tea party either. I'm really at a loss of what to do but I figure i'll end up with 3 in the num. and 1 in the denom since the final answer is 3.

2. ## Re: Limit help, radical in the numerator!!! (Answer provided but idk how it comes abo

Originally Posted by skinsdomination09
the problem is

the lim as x approaches inifinity of

√(9x^6 - 1)/(x^3 - 1)

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

3. ## Re: Limit help, radical in the numerator!!! (Answer provided but idk how it comes abo

Originally Posted by Plato
$\frac{{\sqrt {9{x^6} - 1} }}{{{x^3} - 1}} = \frac{{\sqrt {9 - \dfrac{1}{{{x^6}}}} }}{{1 - \dfrac{1}{{{x^3}}}}}$
I'm sorry I gotta be honest I really don't see what you did there. did you multiply by 1/x^6 or divide by x

4. ## Re: Limit help, radical in the numerator!!! (Answer provided but idk how it comes abo

Originally Posted by skinsdomination09
I'm sorry I gotta be honest I really don't see what you did there. did you multiply by 1/x^6 or divide by x
divide numerator and denominator by $x^3$