Question

Infinite limits question

Answer 1

Is that the question

Answer 2

BRING IT OOOOOOOONNNNNNNNNNN!!!!

Answer 3

\[\frac{ \sqrt{9x^6-x} }{ x^3+1 }\]

Answer 4

x ->-1

Answer 5

?

Answer 6

so i tried multiplying the top and bottom by x^-6

Answer 7

x->inf

Answer 8

square the top and bottom

Answer 9

well you can't do that lol

Answer 10

Sorry but if you could square the top and bottom on this one, you might as well just change it to 1/x

Answer 11

Can't you do that to cancel the square root
Well there is that too

Answer 12

i get a zero on the bottom if i multiply by x^-6

Answer 13

No I was joking, you can't do that.

Answer 14

you could multiply top and bottom by sqrt(9x^6-x)

Answer 15

that would get rid of the square on top and leave you with one on the bottom

Answer 16

ok

Answer 17

can you post the answer? I just want to see if I did it right.

Answer 18

3

Answer 19

yea I did do it right.. but I can't really explain it lol and the teacher would not really accept my explanation.

Answer 20

But here's my attempt at explanation:
sqrt(9x^6 - x)/(x^3+1), basically you would wanna know what approaches infinity faster, top or bottom?

Answer 21

well if the bottom is faster it will be zero

Answer 22

And how would you see that? By "taking the square root", if you rewrote this as:
sqrt(9) sqrt(x^6-x)/(x^3+1)

Answer 23

so you have 3 * sqrt(x^6-x)/(x^3+1)

Answer 24

If you "took" the square root, the highest power on top would be 3, sqrt(x^6) = x^3, and the highest power on the bottom is 3 as well, so basically this would mean that both top and bottom are approaching infinity at the same rate.

Answer 25

still make it a zero wouldnt it?

Answer 26

No, when two things go to infinity at the same rate, they cancel each other and become 1

Answer 27

for example, lim x-> inf (x^3/x^3) = lim x->inf (1) = 1

Answer 28

So since the fraction part cancels each other out (it becomes 1), the only thing that remains is sqrt(9), or 3.

Answer 29

can i somehow factor out an x out of the top and bottom?

Answer 30

You could use the L'Hopital's Rule on this, but it would get messy.

Answer 31

Okay let's do it the long way.. First get top and bottom under a common square root.

Answer 32

\[\frac{ \sqrt{x}\sqrt{9x^5-1} }{ x(x^2+\frac{ 1 }{ x }) }\]

Answer 33

is that possible?

Answer 34

yea but you're making it harder.. here:
So:
\[\frac{\sqrt{9x^6-x}}{x^3+1}\] = \[\sqrt{\frac{9x^6-x}{(x^3+1)^2}}\]
makes sense?

Answer 35

ya, whyd the bottom get squared?

Answer 36

Because now the highest power on the bottom is 6 as well:
\[\sqrt{\frac{9x^6-x}{x^6+2x^3+1}}\]

Answer 37

Now you can pull out the square root to the outside of the limit.. Sorry this may take a while to type out, not that good at latex.

Answer 38

\[\sqrt{\frac{ 9x^6-x }{ x^6+2x^3+1 }}\frac{ x^{-6} }{ x^{-6} }\]

Answer 39

can i do that?

Answer 40

\[\sqrt{\frac{ 9-0 }{ 1+0+0 }}\]

Answer 41

yea

Answer 42

lol you found a shorter way, GOOD JOB!

Answer 43

so, the only thing im having trouble with is the part where you put it all ubder a square root

Answer 44

and squared it

Answer 45

well the bottom did not have a square root right?

Answer 46

right

Answer 47

\[*le \space expression* = \sqrt{(*le \space expression*)^2}\]

Answer 48

ahhh

Answer 49

ok, i will remember that

Answer 50

Please do :D

Answer 51

thanks bah!

Answer 52

2 = sqrt(2^2) = sqrt(4) = 2

Answer 53

np cal!