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anonymous

  • 4 years ago

what is the limit of lim X--->0 sin^2 * 7t / t

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  1. myininaya
    • 4 years ago
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    \[\lim_{t \rightarrow 0}\frac{\sin^2(7t)}{t}=\lim_{t \rightarrow 0}\frac{\sin(7t) \cdot \sin(7t)}{t}\] \[\lim_{t \rightarrow 0}\frac{\sin(7t)}{t} \cdot \sin(7t)=7 \cdot \lim_{t \rightarrow 0}\frac{\sin(7t)}{7t} \cdot \lim_{t \rightarrow 0}\sin(7t)\]

  2. myininaya
    • 4 years ago
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    \[=7(1)(\sin(7 \cdot 0))=7\sin(0)=7(0)=0\]

  3. pokemon23
    • 4 years ago
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    i don't understand high level math

  4. anonymous
    • 4 years ago
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    so its 0?

  5. pokemon23
    • 4 years ago
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    all I understand is sin from triangle

  6. pokemon23
    • 4 years ago
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    yes it 0

  7. myininaya
    • 4 years ago
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    did you really mean x->0?

  8. pokemon23
    • 4 years ago
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    watch the lanaguage

  9. pokemon23
    • 4 years ago
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    read the code of conduct

  10. anonymous
    • 4 years ago
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    lim x→0 x^2 / sin^2 4x

  11. myininaya
    • 4 years ago
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    yeah just to be careful i wouldn't say that so you also meant sin^2 * 7t or sin^2(7t) because sin^2 * 7t doesn't make sense

  12. anonymous
    • 4 years ago
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    i got it right :) and my bad buddy didnt know sucks was a bad word

  13. myininaya
    • 4 years ago
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    so this changes thing since x->0 and not t

  14. pokemon23
    • 4 years ago
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    not that...

  15. pokemon23
    • 4 years ago
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    jsut be careful what you say to people..

  16. myininaya
    • 4 years ago
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    well i think when you put it with that other word its bad

  17. myininaya
    • 4 years ago
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    \[\lim_{x \rightarrow 0}\frac{\sin^2(7t)}{t}\] so this really was the problem now what I thought it was

  18. anonymous
    • 4 years ago
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    thats one i just did was a new one, that i dont get

  19. myininaya
    • 4 years ago
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    \[\lim_{x \rightarrow 0}\frac{\sin^2(7t)}{t}=\frac{\sin^2(7t)}{t}\]

  20. myininaya
    • 4 years ago
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    i thought you meant t->0 earlier

  21. myininaya
    • 4 years ago
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    now that i know you actually meant x->0 this is correct

  22. anonymous
    • 4 years ago
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    the one i did earlier had 7t sin^2.... but i just posted one that was like lim x→0 x^2 / sin^2 4x

  23. myininaya
    • 4 years ago
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    i'm confused

  24. myininaya
    • 4 years ago
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    which one did you mean? x->0 or t->0

  25. anonymous
    • 4 years ago
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    i need help with the last one where3 x -->0

  26. myininaya
    • 4 years ago
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    i been talking about the first one you put up

  27. myininaya
    • 4 years ago
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    t->0 or x->0?

  28. anonymous
    • 4 years ago
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    the first one is right you did great!

  29. myininaya
    • 4 years ago
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    you said x->0 i did the problem for t->0 not x->0

  30. myininaya
    • 4 years ago
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    |dw:1328073227704:dw|

  31. myininaya
    • 4 years ago
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    i did this problem that you posted first fro t->0

  32. myininaya
    • 4 years ago
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    not x->0

  33. myininaya
    • 4 years ago
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    i thought you meant t->0

  34. anonymous
    • 4 years ago
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    i was t on that one

  35. myininaya
    • 4 years ago
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    t->0?

  36. anonymous
    • 4 years ago
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    yeH

  37. myininaya
    • 4 years ago
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    ok! then the first way i did it was correct

  38. anonymous
    • 4 years ago
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    but the new one has x -> 0 x^2/ sin^2 * 4x

  39. myininaya
    • 4 years ago
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    look for 4x/sin(4x) this will go to 1 as x->0

  40. anonymous
    • 4 years ago
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    so its 1?

  41. myininaya
    • 4 years ago
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    4x/(sin(4x))->1 as x->0 so look for 4x/sin(4x)

  42. anonymous
    • 4 years ago
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    im confuse :/

  43. myininaya
    • 4 years ago
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    \[\lim_{x \rightarrow 0}\frac{4x}{\sin(4x)}=1\]

  44. myininaya
    • 4 years ago
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    use this

  45. myininaya
    • 4 years ago
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    \[\lim_{x \rightarrow 0}\frac{x^2}{\sin^2(4x)}=\lim_{x \rightarrow 0}\frac{x}{\sin(4x)} \cdot \frac{x}{\sin(4x)}\]

  46. anonymous
    • 4 years ago
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    so now i solve for x?

  47. myininaya
    • 4 years ago
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    there is no equation we aren't solving for x we are trying to find the limit

  48. myininaya
    • 4 years ago
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    look for what i told you to look for

  49. anonymous
    • 4 years ago
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    so i plug in 1 in x's place

  50. myininaya
    • 4 years ago
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    \[=\lim_{x \rightarrow 0}\frac{4x}{4 \sin(4x)} \cdot \frac{4x}{4 \sin(4x)} \]

  51. myininaya
    • 4 years ago
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    no use what i told you to use

  52. myininaya
    • 4 years ago
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    \[\lim_{x \rightarrow 0}\frac{4x}{\sin(4x)}=1 \] remember this!

  53. anonymous
    • 4 years ago
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    is the answer 1/4

  54. myininaya
    • 4 years ago
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    |dw:1328073897366:dw|

  55. myininaya
    • 4 years ago
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    |dw:1328073931961:dw|

  56. myininaya
    • 4 years ago
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    what do those circled things go to?

  57. anonymous
    • 4 years ago
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    the problem right?, the original one

  58. myininaya
    • 4 years ago
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    \[\lim_{x \rightarrow 0}\frac{4x}{\sin(4x)}=1\]

  59. myininaya
    • 4 years ago
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    so what do those circled things go to?

  60. myininaya
    • 4 years ago
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    |dw:1328074125358:dw|

  61. myininaya
    • 4 years ago
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    they go to 1

  62. myininaya
    • 4 years ago
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    so what do we have ?

  63. anonymous
    • 4 years ago
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    okay, I see that now, so do i add the two fours thats left over together? or just 1 +1

  64. myininaya
    • 4 years ago
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    this is multiplication there is no addition here

  65. anonymous
    • 4 years ago
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    raginsquirrel says its your bday

  66. anonymous
    • 4 years ago
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    so 4 *4

  67. myininaya
    • 4 years ago
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    and both of those 4's are on bottom right?

  68. anonymous
    • 4 years ago
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    1/16

  69. myininaya
    • 4 years ago
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    yes

  70. anonymous
    • 4 years ago
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    wow... youre a genius, my teaher is foreign so i cant understand her, and she doesnt ever explain

  71. anonymous
    • 4 years ago
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    so how about 1- cos9x/ sin 9x bc cos and sine are oppositeees arent they?

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