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\[\lim_{x \rightarrow -\infty} \frac{ x^4*\sin \frac{ 1 }{ x }+x^2 }{ 1+\left| x \right|^3}\]

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Yep.. working on it. Did you try using (a^3 + b^3) identity? not sure whether that will work or not.

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Other answers:

x=-1/h and h->0
on that |x| is that a cube? I can't read for some reason. I can click on the code.
yes it is a cube
I suggest you simplify the denominator... use the identity.
that mod x will open with minus?
may be divide numerator and denominator by x^3
so since x<0 then |x|=-x and since we have |x|^3 then we could replace |x|^3 with (-x)^3=-x^3 now recall that if u->0 then sin(u)/u->1 see if you can use that here divide both top and bottom by 1/x
the some l'hopital could be used :)
for large x \[\Large \frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ 1+\left| x \right|^3}\approx \frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ \left| x \right|^3}\] \[\Large=\frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ x^2\left| x \right|}\] \[\Large=\frac{ x^2\sin \frac{ 1 }{ x }+1 }{ \left| x \right|}\] \[\Large\approx\frac{ x^2\sin \frac{ 1 }{ x } }{ \left| x \right|}=\frac{x}{|x|}x\sin\frac{1}{x}\]
we have to write x=-1/h and then h->0
as \(x\to \infty\) you then get \((-1)\cdot 1=-1\)
x to -infinity you get what i have above
@Zarkon how did u use the approximation in the first step?
in my first step I got rid of the 1 since for large x the 1 really contributes almost nothing
and by large x I mean large negative (obviously)
I would have done it like this: (but this is because i'm not commander data like zarkon (who knows all because he is a superior being to a human) \[\lim_{x \rightarrow - \infty}\frac{\frac{x^4 \sin(\frac{1}{x})}{\frac{1}{x}}+\frac{x^2}{\frac{1}{x}}}{\frac{1}{\frac{1}{x}}-\frac{x^3}{\frac{1}{x}}}\] then use the fact that if u->0 then sin(u)/u->1 it should be pretty easy after this point though.
as xtends to - infinity mod(x)=-x now try to put \[x=\frac{ -1 }{ t}\]
i was thinking that only
or i guess you could have just said \[\lim_{x \rightarrow -\infty}\frac{x^4 \sin(\frac{1}{x})+x^2}{1-x^3}=\lim_{x \rightarrow -\infty}\frac{x^3 \frac{\sin(\frac{1}{x})}{\frac{1}{x}}+x^2}{1-x^3}\]
do u get it ? @cody_123
can u write the intermediate steps too? thanks
|dw:1402332508279:dw| now just substitute x=(-1/t)
\[\lim_{x \rightarrow -\infty} \frac{ (\frac{ -1 }{ t })^4*\sin(-t)+(\frac{ -1 }{ t^2 }) }{1+ \frac{-1 }{ t^3 } }\]
sorry t->0
is it correct @sidsiddhartha ?
no x=-1/t so |dw:1402332969397:dw| it will be like this
ok @sidsiddhartha then ?
now just try to simplify it little more
\[\frac{ \frac{ -1 }{ t }*\sin t+t}{ 1+t^3 }\]
ya good now can do it :)
w8 a second in denominator i should be 1-t^3
nope its all right now just use limit t tends to 0 (sint/t)=1
\[1+\frac{ -1 }{ t^3 }=\frac{ t^3-1 }{ t^3 }\]
so t^3-1 will come in denominator?
|dw:1402334075049:dw| its not "-"
oh silly mistake :P
\[\frac{ \frac{ -1 }{ t }*\sin t+t }{ 1+t^2 }\]
it should be (t^3+1)
ya in the denominator
\[\frac{ t-1 }{ t^3+1 }\]
yeah :)

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