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anonymous

  • 5 years ago

Evaluate the following limit: lim x->0 sin(x) cos (x)/x

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  1. anonymous
    • 5 years ago
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    \[\lim_{x \rightarrow 0}sinxcosx/x=\lim_{x \rightarrow 0}sinx/x*\lim_{x \rightarrow 0}\cos(x)\] both limits are 1. So the whole limit is 1

  2. anonymous
    • 5 years ago
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    cos x/ x = 1?

  3. myininaya
    • 5 years ago
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    sinx/x->1 cosx is left over but also goes to 1 both as x->0

  4. anonymous
    • 5 years ago
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    i got DNE

  5. anonymous
    • 5 years ago
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    no the limit exists, it is 1. cos(x) goes to 1 and sinx/x goes to 1 as well

  6. anonymous
    • 5 years ago
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    because cos x / x = 1/0... no?

  7. myininaya
    • 5 years ago
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    lol can you not write the function as sinx/x *cosx?

  8. anonymous
    • 5 years ago
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    you aren't plugging in 0 though it's a limit

  9. anonymous
    • 5 years ago
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    there are limits that have a division by 0 and still exist it is about infinitesimal behavior

  10. anonymous
    • 5 years ago
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    no i wrote sinx / x * cosx / x

  11. anonymous
    • 5 years ago
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    it is not x^2 on the bottom

  12. myininaya
    • 5 years ago
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    (5*6)/(5)=6 but by what you say we can do or suppose to do 5/5*6/5=6/5

  13. myininaya
    • 5 years ago
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    i mean (5*6)/5=5/5*6=1*6=6 and you say (5*6)/5=5/5*6/5=6/5 is that what you are saying this second line right here?

  14. anonymous
    • 5 years ago
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    i seperate the fraction

  15. myininaya
    • 5 years ago
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    i think you are thinking about addition on fractions (a+b)/c=a/c+b/c

  16. anonymous
    • 5 years ago
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    sorry guys the lag go the best of me

  17. anonymous
    • 5 years ago
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    no i mean the 2nd line what you di

  18. anonymous
    • 5 years ago
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    did*

  19. myininaya
    • 5 years ago
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    you can't do (5*6)/5=5/5*6/5 but you do (5*6)/5=5/5*6 or 5/6*5=6

  20. anonymous
    • 5 years ago
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    yes the 2nd line

  21. anonymous
    • 5 years ago
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    yes the 2nd line

  22. myininaya
    • 5 years ago
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    so we can write sinx/x*cosx?

  23. anonymous
    • 5 years ago
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    no sin x/x * cox x /x

  24. myininaya
    • 5 years ago
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    but you just said we can't do (5*6)/5=5/5*6/5

  25. myininaya
    • 5 years ago
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    (sinx*cosx)/x=sinx/x * cosx

  26. myininaya
    • 5 years ago
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    or =cosx/x * sinx ( it does us no good to write it like this though)

  27. anonymous
    • 5 years ago
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    so how do i work it? i am confused

  28. myininaya
    • 5 years ago
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    write (sinx*cosx)/x=sinx/x*cosx where does sinx/x go to as x gets closer to 1?

  29. myininaya
    • 5 years ago
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    as x gets closer to 0?*

  30. anonymous
    • 5 years ago
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    do you know how to use latex here because i dont understand what you typed

  31. anonymous
    • 5 years ago
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    so the answer is 1 i dont undertstand why it cant be done the way i did it i understand what you typed

  32. myininaya
    • 5 years ago
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    so you still don't understand why we can't do: \[\frac{\sin(x)\cos(x)}{x}=\frac{\sin(x)}{x}\frac{\cos(x)}{x}?\]

  33. myininaya
    • 5 years ago
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    we don't have x^2 on bottom? we only have x

  34. anonymous
    • 5 years ago
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    yes \\

  35. myininaya
    • 5 years ago
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    \[\frac{5*6}{5}\neq \frac{5}{5}*\frac{6}{5}\]

  36. myininaya
    • 5 years ago
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    do you agree with this statement?

  37. anonymous
    • 5 years ago
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    ok...

  38. myininaya
    • 5 years ago
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    \[\frac{5*6}{5}=\frac{5}{5}*6=6\]

  39. myininaya
    • 5 years ago
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    \[\lim_{x \rightarrow 0} \frac{\sin(x)\cos(x)}{x}=\lim_{x \rightarrow 0}\frac{\sin(x)}{x}*\cos(x)\]

  40. anonymous
    • 5 years ago
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    k

  41. myininaya
    • 5 years ago
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    \[=\lim_{x \rightarrow 0}\frac{\sin(x)}{x}*\lim_{x \rightarrow 0}\cos(x)=1*1=1\]

  42. anonymous
    • 5 years ago
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    i guess i gotta recheck my fractions thank you so much i have 1 more question if your not busy can you help me with that please

  43. myininaya
    • 5 years ago
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    i gotta pee i might return

  44. anonymous
    • 5 years ago
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    lol ok THANK YOU SO MUCH

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