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ujuge321

  • 3 years ago

calculate the following limits: lim sin(2x)/sin(3x) as x approaches to 0.

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  1. Outkast3r09
    • 3 years ago
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    use lhopitals rule which is \[\frac{\frac{d}{dx}[\sin(2x)]}{\frac{d}{dx}[\sin(3x)]}\]

  2. Outkast3r09
    • 3 years ago
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    when you do that you will simply get 2/3

  3. Outkast3r09
    • 3 years ago
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    so your limit will be 2/3

  4. ujuge321
    • 3 years ago
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    show me the process. i never learned that in class.

  5. Outkast3r09
    • 3 years ago
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    what calculus class are you in?

  6. raheen
    • 3 years ago
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    using le'hopital rule ==> lim 3co(2x)/3cos(3x) = 2/3

  7. ujuge321
    • 3 years ago
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    calculus 1. Thanks.

  8. raheen
    • 3 years ago
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    using le'hopital rule ==> lim 3cos(2x)/3cos(3x) = 2/3

  9. JamesJ
    • 3 years ago
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    If you don't yet know l'Hopitals rule, I assume you do know already that the limit as x --> 0 of (sin x) / x is 1. Now write\[\frac{\sin 2x }{\sin 3x} = \frac{2}{3} \frac{\sin 2x/(2x)}{\sin 3x/(3x)}\] Then you observe that the limits of the ratios of sin over x are all 1 and hence the answer is 2/3.

  10. ujuge321
    • 3 years ago
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    thank you,

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