cody_123
\[\lim_{x \rightarrow -\infty} \frac{ x^4*\sin \frac{ 1 }{ x }+x^2 }{ 1+\left| x \right|^3}\]
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cody_123
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@ganeshie8 ?
cody_123
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@mathslover ?
mathslover
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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.
cody_123
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x=-1/h and h->0
myininaya
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on that |x| is that a cube?
I can't read for some reason.
I can click on the code.
cody_123
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yes it is a cube
mathslover
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I suggest you simplify the denominator... use the identity.
cody_123
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that mod x will open with minus?
cody_123
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?
ganeshie8
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may be divide numerator and denominator by x^3
myininaya
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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
myininaya
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the some l'hopital could be used :)
Zarkon
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for large x
\[\Large \frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ 1+\left| x \right|^3}\approx \frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ \left| x \right|^3}\]
\[\Large=\frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ x^2\left| x \right|}\]
\[\Large=\frac{ x^2\sin \frac{ 1 }{ x }+1 }{ \left| x \right|}\]
\[\Large\approx\frac{ x^2\sin \frac{ 1 }{ x } }{ \left| x \right|}=\frac{x}{|x|}x\sin\frac{1}{x}\]
cody_123
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we have to write x=-1/h and then h->0
Zarkon
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as \(x\to \infty\) you then get \((-1)\cdot 1=-1\)
Zarkon
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x to -infinity you get what i have above
cody_123
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@Zarkon how did u use the approximation in the first step?
cody_123
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@ganeshie8 ?
cody_123
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@mathslover ?
Zarkon
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in my first step I got rid of the 1 since for large x the 1 really contributes almost nothing
Zarkon
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and by large x I mean large negative (obviously)
cody_123
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@mathslover ?
myininaya
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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.
sidsiddhartha
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as xtends to - infinity mod(x)=-x
now try to put
\[x=\frac{ -1 }{ t}\]
cody_123
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ya @sidsiddhartha
cody_123
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i was thinking that only
myininaya
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or i guess you could have just said
\[\lim_{x \rightarrow -\infty}\frac{x^4 \sin(\frac{1}{x})+x^2}{1-x^3}=\lim_{x \rightarrow -\infty}\frac{x^3 \frac{\sin(\frac{1}{x})}{\frac{1}{x}}+x^2}{1-x^3}\]
sidsiddhartha
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|dw:1402332193100:dw|
sidsiddhartha
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do u get it ? @cody_123
cody_123
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can u write the intermediate steps too? thanks
sidsiddhartha
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|dw:1402332508279:dw|
now just
substitute
x=(-1/t)
cody_123
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\[\lim_{x \rightarrow -\infty} \frac{ (\frac{ -1 }{ t })^4*\sin(-t)+(\frac{ -1 }{ t^2 }) }{1+ \frac{-1 }{ t^3 } }\]
cody_123
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sorry t->0
cody_123
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@sidsiddhartha ?
cody_123
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is it correct @sidsiddhartha ?
sidsiddhartha
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no x=-1/t
so
|dw:1402332969397:dw|
it will be like this
cody_123
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ok @sidsiddhartha then ?
sidsiddhartha
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now just try to simplify it little more
cody_123
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\[\frac{ \frac{ -1 }{ t }*\sin t+t}{ 1+t^3 }\]
cody_123
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@sidsiddhartha ?
sidsiddhartha
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ya good now can do it :)
cody_123
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w8 a second in denominator i should be 1-t^3
cody_123
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@sidsiddhartha ?
sidsiddhartha
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nope its all right
now just use limit t tends to 0 (sint/t)=1
cody_123
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@sidsiddhartha
cody_123
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\[1+\frac{ -1 }{ t^3 }=\frac{ t^3-1 }{ t^3 }\]
cody_123
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so t^3-1 will come in denominator?
cody_123
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@sidsiddhartha ?
sidsiddhartha
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|dw:1402334075049:dw|
its not "-"
cody_123
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oh silly mistake :P
cody_123
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\[\frac{ \frac{ -1 }{ t }*\sin t+t }{ 1+t^2 }\]
cody_123
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@sidsiddhartha ?
sidsiddhartha
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it should be (t^3+1)
cody_123
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where?
cody_123
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ya in the denominator
cody_123
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\[\frac{ t-1 }{ t^3+1 }\]
cody_123
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-1
sidsiddhartha
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yeah :)