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cody_123
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\[\lim_{x \rightarrow \infty} \frac{ x^4*\sin \frac{ 1 }{ x }+x^2 }{ 1+\left x \right^3}\]
 one month ago
 one month ago
cody_123 Group Title
\[\lim_{x \rightarrow \infty} \frac{ x^4*\sin \frac{ 1 }{ x }+x^2 }{ 1+\left x \right^3}\]
 one month ago
 one month ago

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cody_123 Group TitleBest ResponseYou've already chosen the best response.0
@ganeshie8 ?
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
@mathslover ?
 one month ago

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

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
x=1/h and h>0
 one month ago

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

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
yes it is a cube
 one month ago

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

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
that mod x will open with minus?
 one month ago

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

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
the some l'hopital could be used :)
 one month ago

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

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
we have to write x=1/h and then h>0
 one month ago

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

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

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
@Zarkon how did u use the approximation in the first step?
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
@ganeshie8 ?
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
@mathslover ?
 one month ago

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

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

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
@mathslover ?
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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.
 one month ago

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

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
ya @sidsiddhartha
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
i was thinking that only
 one month ago

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

sidsiddhartha Group TitleBest ResponseYou've already chosen the best response.3
dw:1402332193100:dw
 one month ago

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

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
can u write the intermediate steps too? thanks
 one month ago

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

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
\[\lim_{x \rightarrow \infty} \frac{ (\frac{ 1 }{ t })^4*\sin(t)+(\frac{ 1 }{ t^2 }) }{1+ \frac{1 }{ t^3 } }\]
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
sorry t>0
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
@sidsiddhartha ?
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
is it correct @sidsiddhartha ?
 one month ago

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

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
ok @sidsiddhartha then ?
 one month ago

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

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ \frac{ 1 }{ t }*\sin t+t}{ 1+t^3 }\]
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
@sidsiddhartha ?
 one month ago

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

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
w8 a second in denominator i should be 1t^3
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
@sidsiddhartha ?
 one month ago

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

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
@sidsiddhartha
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
\[1+\frac{ 1 }{ t^3 }=\frac{ t^31 }{ t^3 }\]
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
so t^31 will come in denominator?
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
@sidsiddhartha ?
 one month ago

sidsiddhartha Group TitleBest ResponseYou've already chosen the best response.3
dw:1402334075049:dw its not ""
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
oh silly mistake :P
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ \frac{ 1 }{ t }*\sin t+t }{ 1+t^2 }\]
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
@sidsiddhartha ?
 one month ago

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

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
ya in the denominator
 one month ago

cody_123 Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ t1 }{ t^3+1 }\]
 one month ago

sidsiddhartha Group TitleBest ResponseYou've already chosen the best response.3
yeah :)
 one month ago
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