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hba

  • 3 years ago

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  1. hba
    • 3 years ago
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    \[\lim_{x \rightarrow 0}\frac{ \sin2x }{ \sin3x }\]

  2. math_proof
    • 3 years ago
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    2/3

  3. hba
    • 3 years ago
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    Please Explain :(

  4. math_proof
    • 3 years ago
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    ok

  5. math_proof
    • 3 years ago
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    one min

  6. math_proof
    • 3 years ago
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    since you have indeterminate form, you have to use L'hospital rule

  7. math_proof
    • 3 years ago
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    then when you do derivative you get \[\frac{ 2\cos2x }{ 3\cos3x }\]

  8. hba
    • 3 years ago
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    I dont know l hospitals rule.

  9. math_proof
    • 3 years ago
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    you don't?

  10. math_proof
    • 3 years ago
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    you know like squeeze theorem?

  11. hba
    • 3 years ago
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    Please teach me how to solve it,i am helpless :(

  12. math_proof
    • 3 years ago
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    oki so there are couple cases when you have indeterminate form, for example 0/0 , inf/inf

  13. math_proof
    • 3 years ago
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    so what you do then is basically take the derivative of the top and the bottom

  14. math_proof
    • 3 years ago
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    for example \[\lim_{x \rightarrow 1} \frac{ x^3+x^2=2x }{ x-1 }\]

  15. math_proof
    • 3 years ago
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    it suppose to be x^3+x^2-2x

  16. math_proof
    • 3 years ago
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    so if you plug 1 in you get 0/0 which is indeterminate as it has no meaning. so you have to apply l'hospital, derivative of the top and bottom

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