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swin2013

  • 2 years ago

Determine the limit

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  1. swin2013
    • 2 years ago
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    \[\lim x-> 0 \frac{ 3\sin4x }{ \sin3x }\]

  2. swin2013
    • 2 years ago
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    sorry that's limit approaching zero lol

  3. hartnn
    • 2 years ago
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    so do u know the basic limit of sin x/x =... ?

  4. swin2013
    • 2 years ago
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    yes it's = 1?

  5. hartnn
    • 2 years ago
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    so, according to that formula , u need the denominator same as angle of sin. so, multiply and divide by 4x in numerator and multiply and divide by 3x in denominator

  6. hartnn
    • 2 years ago
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    \(\huge \lim \limits_{x-> 0} \frac{ 3 \frac{4x.\sin4x}{4x} }{ 3x.\frac{\sin3x}{3x} }\)

  7. zepdrix
    • 2 years ago
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    That seems like a bit of overkill hart :o don't you just need 4x/4x?

  8. hartnn
    • 2 years ago
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    yeah, thats correct, but i wanted to brig out the point that whenever you see sin ax , multiply and divide by ax

  9. swin2013
    • 2 years ago
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    well can i bring out what's in the denominator first?

  10. swin2013
    • 2 years ago
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    by multiplying the denominator by 3?

  11. zepdrix
    • 2 years ago
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    \[\large \lim_{x \rightarrow 0}\frac{4x}{4x}\frac{3\sin4x}{\sin3x} \quad = \quad 4\lim_{x \rightarrow 0}\frac{\sin4x}{4x}\frac{3x}{\sin3x}\]

  12. zepdrix
    • 2 years ago
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    Oh ok i see what you meant :)

  13. zepdrix
    • 2 years ago
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    The upside down one will also give you 1 swin.

  14. zepdrix
    • 2 years ago
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    It might make more sense if you look at the way hart wrote it out.

  15. hartnn
    • 2 years ago
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    yeah, lim x->0 x/sin x also equals 1

  16. swin2013
    • 2 years ago
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    chyea I'm trying to find a quicker way to solve for my final hahaha! i initially split them up so it was 3sin4x/1 * 1/sin3x. I'm not sure if that's legal

  17. hartnn
    • 2 years ago
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    yeah, its legal step, but you will still need the form sin x/x so you need to multiply and divide by the angle of sin

  18. swin2013
    • 2 years ago
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    So it will be 4?

  19. zepdrix
    • 2 years ago
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    \[\large \color{green}{Good}\quad \color{blue}{job!}\]

  20. swin2013
    • 2 years ago
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    Lol thanks guys!! A couple of hours till I take my final :)

  21. zepdrix
    • 2 years ago
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    Oooo exciting! :O keep cramming. Sleep is good also.

  22. hartnn
    • 2 years ago
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    do u need to show steps ?

  23. hartnn
    • 2 years ago
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    because if you do, while taking the limit, you also need to show that 3x->0 and 4x-> 0

  24. swin2013
    • 2 years ago
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    Probably not, because I think it's just simplifying it to make sinx/x = 1. But I understand how you solved it, Thanks for the review :)

  25. hartnn
    • 2 years ago
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    ok, welcome ^_^

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