Littlebird
  • Littlebird
lim x-> -infinity sqrt(4x^6-x)/(x^3+5)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Littlebird
  • Littlebird
@DanJS
Littlebird
  • Littlebird
This square root is messing me up.
DanJS
  • DanJS
square root of that ratio, or just root of numerator

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More answers

Littlebird
  • Littlebird
of the numerator
DanJS
  • DanJS
similar to tha tlast one, i think , ...
DanJS
  • DanJS
take as much of the x terms out of the root as you can
DanJS
  • 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
  • Littlebird
|dw:1443228681001:dw|
Littlebird
  • 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
  • Littlebird
I also tried conjugating it, but that also looked insane
DanJS
  • DanJS
you can't take the root of a quantity with addition like that
DanJS
  • DanJS
root(x+a) is not same as root(x) + root (a)
Littlebird
  • Littlebird
I sort of knew that, but I was willing to try mathematical blasphemy to solve this
DanJS
  • DanJS
maybe try dividing through by root(x^6)
DanJS
  • DanJS
the numerator and denominator
DanJS
  • DanJS
\[\Large \frac{ [\sqrt{4x^6-x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{ \sqrt{x^6} } }\]
DanJS
  • DanJS
seems promising
DanJS
  • DanJS
trying to get a constant - a fraction over x in the root
DanJS
  • DanJS
and since it is a book prob, root(x^6) = x^3 yay
DanJS
  • DanJS
k, so try reorganizing things
DanJS
  • DanJS
distribute, combine roots, simplify
DanJS
  • DanJS
you can't just divide the denominator by something, you have to do the same to everything else, the numerator
DanJS
  • DanJS
4/6 not same as 4/(6/a)
DanJS
  • DanJS
have to divide everything
Littlebird
  • Littlebird
do I conjugate?
DanJS
  • 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
  • DanJS
\[\sqrt{\frac{ a }{ b }}= \frac{ \sqrt{a} }{ \sqrt{b} }\]
DanJS
  • DanJS
For the top, and for the bottom just distribute and divide if you can
Littlebird
  • Littlebird
i now think I did the division wrong earlier
DanJS
  • 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
  • DanJS
let me know if you dont see how everything worked
DanJS
  • DanJS
\[\large = \frac{ \sqrt{4-\frac{ 1 }{ x^5 }} }{ 1+ \frac{ 5 }{ x^3 } }\]
Littlebird
  • 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
  • DanJS
or x^3, they are the same
DanJS
  • DanJS
root of x^6 is x^3
DanJS
  • DanJS
nice book prob;lem.. hah
DanJS
  • DanJS
\[\large \sqrt{x^6} = [x^6]^{1/2} = x^{6/2} = x^3\]
DanJS
  • DanJS
yeah, was trying to find something to divide by that would leave a number + a fraction over x under the root
DanJS
  • DanJS
after this is done, i think i remember an even better way when the limt is to infinity..
Littlebird
  • Littlebird
|dw:1443231058854:dw|
Littlebird
  • Littlebird
I know I'm missing something, since the answer is negative for some reason
Littlebird
  • 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
  • DanJS
umm, sorry was afk
DanJS
  • DanJS
bad though, forget that
Littlebird
  • Littlebird
ok
DanJS
  • DanJS
property 6 on that pdf
DanJS
  • DanJS
then property 2, it is good
Littlebird
  • Littlebird
where did the negative sign come from?
DanJS
  • DanJS
oh i think i see how it will be negative
DanJS
  • 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
  • DanJS
something to do with an odd power and or negative infinity value, and that root property, lol i can't think
Littlebird
  • Littlebird
I will try to Google your words
DanJS
  • 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
  • DanJS
or probably something simple i just cant think of to fix this as is
DanJS
  • DanJS
since the limit is to negative infinity, you can assume the x value is negative...
DanJS
  • 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
  • Littlebird
I saw it. It looks like I have another obscure rule to remember
DanJS
  • 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
  • DanJS
neg infinity limit
DanJS
  • 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
  • DanJS
@jim_thompson5910
DanJS
  • 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
  • DanJS
than the other parts of the quantities
Littlebird
  • Littlebird
That does work, but I still have to remember the negative
DanJS
  • DanJS
\[\large \lim_{x \rightarrow -\infty} [\frac{ \sqrt{4x^6} }{ x^3 }]\]
DanJS
  • DanJS
i see, that works using the abs value function...
DanJS
  • DanJS
\[\frac{ 2\sqrt{x^6} }{ x^3 }\]
DanJS
  • DanJS
\[\large \frac{ 2\sqrt{[x^3]^2} }{ x^3 }\]
Littlebird
  • Littlebird
Thanks for the help! If I stop replying, it's because of a magic show in 10 min
DanJS
  • DanJS
i think that works, use the absolute value function, the sign should change
Littlebird
  • Littlebird
I will remeber both ways to be safe
DanJS
  • DanJS
lol , calc is 99.9% algebra with a small initial shortcut from calc
DanJS
  • DanJS
\[\large \frac{ 2*\left| x \right|^3 }{ x^3 } \]
DanJS
  • 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
  • DanJS
\[\huge \frac{ 2*(-x)^3 }{ x^3 }\]
DanJS
  • 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
  • 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
  • DanJS
figured what to do, same reason as the solution above, just use the absolute value function
DanJS
  • 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
  • DanJS
\[\large =\frac{ \sqrt{4-1/x^5} }{ \frac{ x^3 }{ \left| x^3 \right| } +\frac{ 5 }{ \left| x^3 \right| }}\]
DanJS
  • 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
  • DanJS
limit value at -infinity = 2 / -1 = - 2

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