Thread: Limit help, radical in the numerator!!! (Answer provided but idk how it comes about)

1. Limit help, radical in the numerator!!! (Answer provided but idk how it comes about)

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)

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

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

Originally Posted by Plato
$\displaystyle \frac{{\sqrt {9{x^6} - 1} }}{{{x^3} - 1}} = \frac{{\sqrt {9 - \dfrac{1}{{{x^6}}}} }}{{1 - \dfrac{1}{{{x^3}}}}}$
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 $\displaystyle x^3$