## cody_123 one year ago $\lim_{x \rightarrow -\infty} \frac{ x^4*\sin \frac{ 1 }{ x }+x^2 }{ 1+\left| x \right|^3}$

1. cody_123

@ganeshie8 ?

2. cody_123

@mathslover ?

3. mathslover

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

4. cody_123

x=-1/h and h->0

5. myininaya

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

6. cody_123

yes it is a cube

7. mathslover

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

8. cody_123

that mod x will open with minus?

9. cody_123

?

10. ganeshie8

may be divide numerator and denominator by x^3

11. myininaya

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

12. myininaya

the some l'hopital could be used :)

13. Zarkon

for large x $\Large \frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ 1+\left| x \right|^3}\approx \frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ \left| x \right|^3}$ $\Large=\frac{ x^4\sin \frac{ 1 }{ x }+x^2 }{ x^2\left| x \right|}$ $\Large=\frac{ x^2\sin \frac{ 1 }{ x }+1 }{ \left| x \right|}$ $\Large\approx\frac{ x^2\sin \frac{ 1 }{ x } }{ \left| x \right|}=\frac{x}{|x|}x\sin\frac{1}{x}$

14. cody_123

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

15. Zarkon

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

16. Zarkon

x to -infinity you get what i have above

17. cody_123

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

18. cody_123

@ganeshie8 ?

19. cody_123

@mathslover ?

20. Zarkon

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

21. Zarkon

and by large x I mean large negative (obviously)

22. cody_123

@mathslover ?

23. myininaya

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.

24. sidsiddhartha

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

25. cody_123

ya @sidsiddhartha

26. cody_123

i was thinking that only

27. myininaya

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

28. sidsiddhartha

|dw:1402332193100:dw|

29. sidsiddhartha

do u get it ? @cody_123

30. cody_123

can u write the intermediate steps too? thanks

31. sidsiddhartha

|dw:1402332508279:dw| now just substitute x=(-1/t)

32. cody_123

$\lim_{x \rightarrow -\infty} \frac{ (\frac{ -1 }{ t })^4*\sin(-t)+(\frac{ -1 }{ t^2 }) }{1+ \frac{-1 }{ t^3 } }$

33. cody_123

sorry t->0

34. cody_123

@sidsiddhartha ?

35. cody_123

is it correct @sidsiddhartha ?

36. sidsiddhartha

no x=-1/t so |dw:1402332969397:dw| it will be like this

37. cody_123

ok @sidsiddhartha then ?

38. sidsiddhartha

now just try to simplify it little more

39. cody_123

$\frac{ \frac{ -1 }{ t }*\sin t+t}{ 1+t^3 }$

40. cody_123

@sidsiddhartha ?

41. sidsiddhartha

ya good now can do it :)

42. cody_123

w8 a second in denominator i should be 1-t^3

43. cody_123

@sidsiddhartha ?

44. sidsiddhartha

nope its all right now just use limit t tends to 0 (sint/t)=1

45. cody_123

@sidsiddhartha

46. cody_123

$1+\frac{ -1 }{ t^3 }=\frac{ t^3-1 }{ t^3 }$

47. cody_123

so t^3-1 will come in denominator?

48. cody_123

@sidsiddhartha ?

49. sidsiddhartha

|dw:1402334075049:dw| its not "-"

50. cody_123

oh silly mistake :P

51. cody_123

$\frac{ \frac{ -1 }{ t }*\sin t+t }{ 1+t^2 }$

52. cody_123

@sidsiddhartha ?

53. sidsiddhartha

it should be (t^3+1)

54. cody_123

where?

55. cody_123

ya in the denominator

56. cody_123

$\frac{ t-1 }{ t^3+1 }$

57. cody_123

-1

58. sidsiddhartha

yeah :)