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anonymous
 one year ago
\[\lim_{x \rightarrow \infty} \frac{ x^4*\sin \frac{ 1 }{ x }+x^2 }{ 1+\left x \right^3}\]
anonymous
 one year ago
\[\lim_{x \rightarrow \infty} \frac{ x^4*\sin \frac{ 1 }{ x }+x^2 }{ 1+\left x \right^3}\]

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mathslover
 one year ago
Best ResponseYou've already chosen the best response.0Yep.. working on it. Did you try using (a^3 + b^3) identity? not sure whether that will work or not.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1on that x is that a cube? I can't read for some reason. I can click on the code.

mathslover
 one year ago
Best ResponseYou've already chosen the best response.0I suggest you simplify the denominator... use the identity.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that mod x will open with minus?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1may be divide numerator and denominator by x^3

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.1the some l'hopital could be used :)

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1for large x \[\Large \frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ 1+\left x \right^3}\approx \frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ \left x \right^3}\] \[\Large=\frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ x^2\left x \right}\] \[\Large=\frac{ x^2\sin \frac{ 1 }{ x }+1 }{ \left x \right}\] \[\Large\approx\frac{ x^2\sin \frac{ 1 }{ x } }{ \left x \right}=\frac{x}{x}x\sin\frac{1}{x}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we have to write x=1/h and then h>0

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1as \(x\to \infty\) you then get \((1)\cdot 1=1\)

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1x to infinity you get what i have above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Zarkon how did u use the approximation in the first step?

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1in my first step I got rid of the 1 since for large x the 1 really contributes almost nothing

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1and by large x I mean large negative (obviously)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.3as xtends to  infinity mod(x)=x now try to put \[x=\frac{ 1 }{ t}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i was thinking that only

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1or i guess you could have just said \[\lim_{x \rightarrow \infty}\frac{x^4 \sin(\frac{1}{x})+x^2}{1x^3}=\lim_{x \rightarrow \infty}\frac{x^3 \frac{\sin(\frac{1}{x})}{\frac{1}{x}}+x^2}{1x^3}\]

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.3dw:1402332193100:dw

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.3do u get it ? @cody_123

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can u write the intermediate steps too? thanks

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.3dw:1402332508279:dw now just substitute x=(1/t)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \infty} \frac{ (\frac{ 1 }{ t })^4*\sin(t)+(\frac{ 1 }{ t^2 }) }{1+ \frac{1 }{ t^3 } }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it correct @sidsiddhartha ?

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.3no x=1/t so dw:1402332969397:dw it will be like this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok @sidsiddhartha then ?

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.3now just try to simplify it little more

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \frac{ 1 }{ t }*\sin t+t}{ 1+t^3 }\]

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.3ya good now can do it :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0w8 a second in denominator i should be 1t^3

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.3nope its all right now just use limit t tends to 0 (sint/t)=1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[1+\frac{ 1 }{ t^3 }=\frac{ t^31 }{ t^3 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so t^31 will come in denominator?

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.3dw:1402334075049:dw its not ""

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \frac{ 1 }{ t }*\sin t+t }{ 1+t^2 }\]

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.3it should be (t^3+1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ya in the denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ t1 }{ t^3+1 }\]
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