Question

Find the limit
lim x approaches infinity (sqrt9x^6-x)/(x^3+9)

Answer 1

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?)

Answer 2

yes

Answer 3

3

Answer 4

with your eyeballs.

Answer 5

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

Answer 6

\[\sqrt{9x^6}=3x^3\]

Answer 7

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

Answer 8

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

Answer 9

\[\frac{3x^3}{x^3}=3\] finished

Answer 10

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

Answer 11

but the textbook showed me a differnt answer

Answer 12

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.

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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