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

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

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

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

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

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

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

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

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

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

myininaya
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0i was thinking that only

myininaya
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.3dw:1402332193100:dw

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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