Question

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

Answer 1

@ganeshie8 ?

Answer 2

@mathslover ?

Answer 3

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

Answer 4

x=-1/h and h->0

Answer 5

on that |x| is that a cube?
I can't read for some reason.
I can click on the code.

Answer 6

yes it is a cube

Answer 7

I suggest you simplify the denominator... use the identity.

Answer 8

that mod x will open with minus?

Answer 9

?

Answer 10

may be divide numerator and denominator by x^3

Answer 11

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

Answer 12

the some l'hopital could be used :)

Answer 13

for large x

\[\Large \frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ 1+\left| x \right|^3}\approx \frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ \left| x \right|^3}\]

\[\Large=\frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ x^2\left| x \right|}\]

\[\Large=\frac{ x^2\sin \frac{ 1 }{ x }+1 }{ \left| x \right|}\]

\[\Large\approx\frac{ x^2\sin \frac{ 1 }{ x } }{ \left| x \right|}=\frac{x}{|x|}x\sin\frac{1}{x}\]

Answer 14

we have to write x=-1/h and then h->0

Answer 15

as \(x\to \infty\) you then get \((-1)\cdot 1=-1\)

Answer 16

x to -infinity you get what i have above

Answer 17

@Zarkon how did u use the approximation in the first step?

Answer 18

@ganeshie8 ?

Answer 19

@mathslover ?

Answer 20

in my first step I got rid of the 1 since for large x the 1 really contributes almost nothing

Answer 21

and by large x I mean large negative (obviously)

Answer 22

@mathslover ?

Answer 23

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.

Answer 24

as xtends to - infinity mod(x)=-x
now try to put
\[x=\frac{ -1 }{ t}\]

Answer 25

ya @sidsiddhartha

Answer 26

i was thinking that only

Answer 27

or i guess you could have just said
\[\lim_{x \rightarrow -\infty}\frac{x^4 \sin(\frac{1}{x})+x^2}{1-x^3}=\lim_{x \rightarrow -\infty}\frac{x^3 \frac{\sin(\frac{1}{x})}{\frac{1}{x}}+x^2}{1-x^3}\]