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Littlebird
 one year ago
lim x> infinity sqrt(4x^6x)/(x^3+5)
Littlebird
 one year ago
lim x> infinity sqrt(4x^6x)/(x^3+5)

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Littlebird
 one year ago
Best ResponseYou've already chosen the best response.0This square root is messing me up.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1square root of that ratio, or just root of numerator

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1similar to tha tlast one, i think , ...

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1take as much of the x terms out of the root as you can

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1try and manipulate that root, so that an infinit x value is in the denominator of a term... so it goes away... maybe

Littlebird
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443228681001:dw

Littlebird
 one year ago
Best ResponseYou've already chosen the best response.0that 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
 one year ago
Best ResponseYou've already chosen the best response.0I also tried conjugating it, but that also looked insane

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1you can't take the root of a quantity with addition like that

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1root(x+a) is not same as root(x) + root (a)

Littlebird
 one year ago
Best ResponseYou've already chosen the best response.0I sort of knew that, but I was willing to try mathematical blasphemy to solve this

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1maybe try dividing through by root(x^6)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1the numerator and denominator

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large \frac{ [\sqrt{4x^6x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{ \sqrt{x^6} } }\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1trying to get a constant  a fraction over x in the root

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1and since it is a book prob, root(x^6) = x^3 yay

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1k, so try reorganizing things

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1distribute, combine roots, simplify

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1you can't just divide the denominator by something, you have to do the same to everything else, the numerator

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1have to divide everything

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1just 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\sqrt{\frac{ a }{ b }}= \frac{ \sqrt{a} }{ \sqrt{b} }\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1For the top, and for the bottom just distribute and divide if you can

Littlebird
 one year ago
Best ResponseYou've already chosen the best response.0i now think I did the division wrong earlier

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large \frac{ [\sqrt{4x^6x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{ \sqrt{x^6} } }~=~\frac{ \sqrt{\frac{ 4x^6x }{ x^6 }} }{ \frac{ x^3 }{ \sqrt{x^6}}+\frac{ 5 }{ \sqrt{x^6} } }\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1let me know if you dont see how everything worked

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\large = \frac{ \sqrt{4\frac{ 1 }{ x^5 }} }{ 1+ \frac{ 5 }{ x^3 } }\]

Littlebird
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1or x^3, they are the same

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1nice book prob;lem.. hah

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\large \sqrt{x^6} = [x^6]^{1/2} = x^{6/2} = x^3\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1yeah, was trying to find something to divide by that would leave a number + a fraction over x under the root

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1after this is done, i think i remember an even better way when the limt is to infinity..

Littlebird
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443231058854:dw

Littlebird
 one year ago
Best ResponseYou've already chosen the best response.0I know I'm missing something, since the answer is negative for some reason

Littlebird
 one year ago
Best ResponseYou've already chosen the best response.0Is 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
 one year ago
Best ResponseYou've already chosen the best response.1then property 2, it is good

Littlebird
 one year ago
Best ResponseYou've already chosen the best response.0where did the negative sign come from?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1oh i think i see how it will be negative

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.1something to do with an odd power and or negative infinity value, and that root property, lol i can't think

Littlebird
 one year ago
Best ResponseYou've already chosen the best response.0I will try to Google your words

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1if you pull out a root x^2 = abs(x) from the initial numorater, that will change the sign i think, just different solving

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1or probably something simple i just cant think of to fix this as is

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1since the limit is to negative infinity, you can assume the x value is negative...

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1look 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
 one year ago
Best ResponseYou've already chosen the best response.0I saw it. It looks like I have another obscure rule to remember

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1if 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\left x \right = x,~~if ~x \ge 0~~and~~~\left x \right=x~~if~x<0\] that is probably where the change in sign comes from

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1oh, 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
 one year ago
Best ResponseYou've already chosen the best response.1than the other parts of the quantities

Littlebird
 one year ago
Best ResponseYou've already chosen the best response.0That does work, but I still have to remember the negative

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\large \lim_{x \rightarrow \infty} [\frac{ \sqrt{4x^6} }{ x^3 }]\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1i see, that works using the abs value function...

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ 2\sqrt{x^6} }{ x^3 }\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\large \frac{ 2\sqrt{[x^3]^2} }{ x^3 }\]

Littlebird
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for the help! If I stop replying, it's because of a magic show in 10 min

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1i think that works, use the absolute value function, the sign should change

Littlebird
 one year ago
Best ResponseYou've already chosen the best response.0I will remeber both ways to be safe

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1lol , calc is 99.9% algebra with a small initial shortcut from calc

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\large \frac{ 2*\left x \right^3 }{ x^3 } \]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1since 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\huge \frac{ 2*(x)^3 }{ x^3 }\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\huge \frac{ 2*(1)^3(x^3 )}{ x^3 } = 2\] yeah, sounds good to me, just wonder what needed to be done, to the first manipulation to put in that sign change...

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large \frac{ [\sqrt{4x^6x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{ \sqrt{x^6} } } = \frac{ \sqrt{4\frac{ 1 }{ x^5 }} }{ 1+ \frac{ 5 }{ x^3 } } \] i am done here for now

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1figured what to do, same reason as the solution above, just use the absolute value function

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large \frac{ [\sqrt{4x^6x}]*\frac{ 1 }{ \sqrt{x^6} } }{[ x^3 + 5]*\frac{ 1 }{ \sqrt{x^6} } }~=~\frac{ \sqrt{\frac{ 4x^6x }{ x^6 }} }{ \frac{ x^3 }{ \sqrt{x^6}}+\frac{ 5 }{ \sqrt{x^6} } }\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\large =\frac{ \sqrt{41/x^5} }{ \frac{ x^3 }{ \left x^3 \right } +\frac{ 5 }{ \left x^3 \right }}\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1same 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
 one year ago
Best ResponseYou've already chosen the best response.1limit value at infinity = 2 / 1 =  2
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