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rainbow22

  • 3 years ago

lim x->0 [(3sin4x)/(sin 3x)] = 4 lim x->0 [(sin 4x/4x)x(3x/sin3x)] How is this true?

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  1. hartnn
    • 3 years ago
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    to get 4x in the denominator, 4x is multiplied in numerator and denominator. got some idea?

  2. Algebraic!
    • 3 years ago
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    Multiply the whole thing by (12x/12x). rearrange.

  3. dumbcow
    • 3 years ago
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    you can also use L'Hopitals Rule -Take derivative of both top and bottom, then reevaluate the limit

  4. rainbow22
    • 3 years ago
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    But it would be undefined if I took the limit of the bottom

  5. Algebraic!
    • 3 years ago
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    Always with l'Hopital's. You'd think it was the only thing anyone ever retained from calc. class.

  6. dumbcow
    • 3 years ago
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    haha..yeah pretty much

  7. rainbow22
    • 3 years ago
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    I haven't learned it yet :(

  8. Algebraic!
    • 3 years ago
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    They're learning about sinc(x) here anyway ... that's a few lectures before the class on l'Hopital's.

  9. hartnn
    • 3 years ago
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    |dw:1347162711646:dw| got it now?

  10. hartnn
    • 3 years ago
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    just use sin t / t when limit t->0 = 1 formula.

  11. Algebraic!
    • 3 years ago
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    @hartnn what did you multiply by? It's cut off in the sketch.

  12. rainbow22
    • 3 years ago
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    Why is it that .... the 4 is moved outside the limit?

  13. hartnn
    • 3 years ago
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    multiplied by 4x in num and denom, and it can be seen,iguess

  14. hartnn
    • 3 years ago
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    the 4 is constant and constants can be moved outside the limit

  15. rainbow22
    • 3 years ago
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    |dw:1347163083569:dw|

  16. rainbow22
    • 3 years ago
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    Sorry, I meant 4x

  17. rainbow22
    • 3 years ago
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    so 12x/sin3x

  18. hartnn
    • 3 years ago
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    now 12x / sin 3x can again be written as 4 * (3x/sin 3x) and using same limit again its 4*1= 4 ok?

  19. Algebraic!
    • 3 years ago
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    |dw:1347163221313:dw|

  20. rainbow22
    • 3 years ago
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    Oh, I get it now! Thank you guys SO MUCH!

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