lim x-> -infinity sqrt(4x^6-x)/(x^3+5)

- Littlebird

lim x-> -infinity sqrt(4x^6-x)/(x^3+5)

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- Littlebird

@DanJS

- Littlebird

This square root is messing me up.

- DanJS

square root of that ratio, or just root of numerator

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## More answers

- Littlebird

of the numerator

- DanJS

similar to tha tlast one, i think , ...

- DanJS

take as much of the x terms out of the root as you can

- DanJS

try and manipulate that root, so that an infinit x value is in the denominator of a term... so it goes away... maybe

- Littlebird

|dw:1443228681001:dw|

- Littlebird

that mess is basically one of my attempts, but it looks like I'm going to get a positive result. wolfram says it is -2, but it won't explain why

- Littlebird

I also tried conjugating it, but that also looked insane

- DanJS

you can't take the root of a quantity with addition like that

- DanJS

root(x+a) is not same as root(x) + root (a)

- Littlebird

I sort of knew that, but I was willing to try mathematical blasphemy to solve this

- DanJS

maybe try dividing through by root(x^6)

- DanJS

the numerator and denominator

- DanJS

\[\Large \frac{ [\sqrt{4x^6-x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{ \sqrt{x^6} } }\]

- DanJS

seems promising

- DanJS

trying to get a constant - a fraction over x in the root

- DanJS

and since it is a book prob, root(x^6) = x^3 yay

- DanJS

k, so try reorganizing things

- DanJS

distribute, combine roots, simplify

- DanJS

you can't just divide the denominator by something, you have to do the same to everything else, the numerator

- DanJS

4/6 not same as 4/(6/a)

- DanJS

have to divide everything

- Littlebird

do I conjugate?

- DanJS

just reorganize to simplify, ill start typing the next piece, remember a root over a root is the same as the root of the fraction together

- DanJS

\[\sqrt{\frac{ a }{ b }}= \frac{ \sqrt{a} }{ \sqrt{b} }\]

- DanJS

For the top, and for the bottom just distribute and divide if you can

- Littlebird

i now think I did the division wrong earlier

- DanJS

\[\Large \frac{ [\sqrt{4x^6-x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{ \sqrt{x^6} } }~=~\frac{ \sqrt{\frac{ 4x^6-x }{ x^6 }} }{ \frac{ x^3 }{ \sqrt{x^6}}+\frac{ 5 }{ \sqrt{x^6} } }\]

- DanJS

let me know if you dont see how everything worked

- DanJS

\[\large = \frac{ \sqrt{4-\frac{ 1 }{ x^5 }} }{ 1+ \frac{ 5 }{ x^3 } }\]

- Littlebird

I feel silly. I just realized now what you meant by dividing by sqrt(x^6). I was still trying to divide by x^3

- DanJS

or x^3, they are the same

- DanJS

root of x^6 is x^3

- DanJS

nice book prob;lem.. hah

- DanJS

\[\large \sqrt{x^6} = [x^6]^{1/2} = x^{6/2} = x^3\]

- DanJS

yeah, was trying to find something to divide by that would leave a number + a fraction over x under the root

- DanJS

after this is done, i think i remember an even better way when the limt is to infinity..

- Littlebird

|dw:1443231058854:dw|

- Littlebird

I know I'm missing something, since the answer is negative for some reason

- Littlebird

Is it because when I divided by sqrt(x^6) I am technically supposed to be dividing by -x^3 since x->-infinity is negative?

- DanJS

umm, sorry was afk

- DanJS

bad though, forget that

- Littlebird

ok

- DanJS

property 6 on that pdf

- DanJS

then property 2, it is good

- Littlebird

where did the negative sign come from?

- DanJS

oh i think i see how it will be negative

- DanJS

i think it has to do with the square root property, a square root is always a positive value
\[\huge \sqrt{[f(x)]^2} = \left| f(x) \right|\]

- DanJS

something to do with an odd power and or negative infinity value, and that root property, lol i can't think

- Littlebird

I will try to Google your words

- DanJS

if you pull out a root x^2 = abs(x) from the initial numorater, that will change the sign i think, just different solving

- DanJS

or probably something simple i just cant think of to fix this as is

- DanJS

since the limit is to negative infinity, you can assume the x value is negative...

- DanJS

look through example 3 on this page, it is similar and probably has the answer in it
http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityI.aspx

- Littlebird

I saw it. It looks like I have another obscure rule to remember

- DanJS

if you have an abs val of x, and the limit is to infinity, then i assume you can call x negative, so the absolute value will be -x

- DanJS

neg infinity limit

- DanJS

\[\left| x \right| = x,~~if ~x \ge 0~~and~~~\left| x \right|=-x~~if~x<0\]
that is probably where the change in sign comes from

- DanJS

@jim_thompson5910

- DanJS

oh, the other thing i think you can do is just forget about the x in the top and the 5 in the bottom , since they are magnitudes less effective for large x values to infinity

- DanJS

than the other parts of the quantities

- Littlebird

That does work, but I still have to remember the negative

- DanJS

\[\large \lim_{x \rightarrow -\infty} [\frac{ \sqrt{4x^6} }{ x^3 }]\]

- DanJS

i see, that works using the abs value function...

- DanJS

\[\frac{ 2\sqrt{x^6} }{ x^3 }\]

- DanJS

\[\large \frac{ 2\sqrt{[x^3]^2} }{ x^3 }\]

- Littlebird

Thanks for the help!
If I stop replying, it's because of a magic show in 10 min

- DanJS

i think that works, use the absolute value function, the sign should change

- Littlebird

I will remeber both ways to be safe

- DanJS

lol , calc is 99.9% algebra with a small initial shortcut from calc

- DanJS

\[\large \frac{ 2*\left| x \right|^3 }{ x^3 } \]

- DanJS

since the limit is out to negative infinity, you can assume that the x value will be negative
From, the absolute value function, \[\left| x \right| = -x~~~if~~~x<0\]

- DanJS

\[\huge \frac{ 2*(-x)^3 }{ x^3 }\]

- DanJS

\[\huge \frac{ 2*(-1)^3(x^3 )}{ x^3 } = -2\]
yeah, sounds good to me, just wonder what needed to be done, to the first manipulation to put in that sign change...

- DanJS

\[\Large \frac{ [\sqrt{4x^6-x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{
\sqrt{x^6} } } = \frac{ \sqrt{4-\frac{ 1 }{ x^5 }} }{ 1+ \frac{ 5 }{ x^3 } } \]
i am done here for now

- DanJS

figured what to do,
same reason as the solution above, just use the absolute value function

- DanJS

- DanJS

\[\large =\frac{ \sqrt{4-1/x^5} }{ \frac{ x^3 }{ \left| x^3 \right| } +\frac{ 5 }{ \left| x^3 \right| }}\]

- DanJS

same logic as the solution above, limit to negative infinity means a negative x value, so the absolute value (x) is -x, and it stays negative for an odd power of 3, there is the change in friggin sign,

- DanJS

limit value at -infinity
= 2 / -1 = - 2

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