## anonymous 5 years ago Find the limit lim x approaches infinity (sqrt9x^6-x)/(x^3+9)

1. mathteacher1729

do you mean $\lim_{x\to \infty} \frac{\sqrt{9x^6-x}}{x^3+9}$ ? (Is that the placement of the square root and the stuff under it?)

2. anonymous

yes

3. anonymous

3

4. anonymous

5. anonymous

here the numerator is not a polynomial, but think of it as one.

6. anonymous

$\sqrt{9x^6}=3x^3$

7. anonymous

so like you said earlier that the bigger the coeffienct is that the answer. so $\sqrt{9x^6}= 3$

8. anonymous

now as we just saw you can ignore the lower stuff.

9. anonymous

$\frac{3x^3}{x^3}=3$ finished

10. anonymous

think of it as the rules we just wrote. even though this is not one polynomial over another you can pretend it is. then the degrees are the same and 3/1=3

11. anonymous

but the textbook showed me a differnt answer

12. mathteacher1729

Start by dividing everything in the numerator and everything in the denominator by x^3. This gives: $\lim x \to \infty \frac{\frac{1}{x^3}\sqrt{9x^6-x}}{\frac{1}{x^3}(x^3+9)}$ Distribute the x^3 appropriately and we have... $\lim_{x \to \infty} \frac{\sqrt{\frac{9x^6}{x^6}-\frac{x}{x^3}}}{(\frac{x^3}{x^3}+\frac{9}{x^3})}$ and this simplifies (thank heavens) to : $\lim_{x \to \infty} \frac{\sqrt{9-\frac{1}{x^2}}}{1+\frac{9}{x^3}}$ As x goes to infinity this limit becomes: $\frac{\sqrt{9-0}}{1}=3$ And that's your answer. Wooo.

13. anonymous

better not have showed a different answer. maybe a different method, but not a different answer.

14. anonymous

mathteacher wrote out all the details, but if i was doing it on an exam i would use my eyes

15. anonymous

ok i will continut to show my professor this step to see if i can get credits thanks

16. anonymous

good idea. that is the "proof" so write that out.