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kvdop

  • 3 years ago

Find the lim of tan(2x)/(3x+sinx) as x approaches 0

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  1. NoelGreco
    • 3 years ago
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    I don't see any quick way without l'Hopital.

  2. pasta
    • 3 years ago
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    l'Hopitals has taken me back to the start

  3. 2le
    • 3 years ago
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    If l'Hopital doesn't work the first time but gives you the condition for it again, you can repeat it. You might have better luck the second time around.

  4. pasta
    • 3 years ago
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    1/4

  5. skywalker94
    • 3 years ago
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    One way is to use l'Hopital, but a faster way is to use the linear approximations for sine and tangent. The answer is 1/2.

  6. pasta
    • 3 years ago
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    THE DIFFERENTIAL IS SEC^2(X)/3+COS(x).numerator is 1 and denominator is 4 answer is 1/4 how did you get 1/2??????

  7. MathPhysics
    • 3 years ago
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    pasta! skywalker94 is right! Using l'Hopital, we'll get 1/2. \[\lim_{x \rightarrow 0}(\tan2x/(3x+sinx))=\lim_{x \rightarrow 0}[2(1+\tan^{2}2x)/(3+cosx)]=1/2\]

  8. adi171
    • 3 years ago
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    use expansion method it is the fastest.... tanx = x + x^3/3............sinx = x - x^3/3!

  9. adi171
    • 3 years ago
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    and from that the ans is 1/2 Math Physics and skywalker ur right..!!!!

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