anonymous
  • anonymous
Find the lim of tan(2x)/(3x+sinx) as x approaches 0
MIT 18.01 Single Variable Calculus (OCW)
schrodinger
  • schrodinger
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NoelGreco
  • NoelGreco
I don't see any quick way without l'Hopital.
anonymous
  • anonymous
l'Hopitals has taken me back to the start
anonymous
  • anonymous
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.

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anonymous
  • anonymous
1/4
anonymous
  • anonymous
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.
anonymous
  • anonymous
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??????
anonymous
  • anonymous
pasta! skywalker94 is right! Using l'Hopital, we'll get 1/2. \[\lim_{x \rightarrow 0}(\tan2x/(3x+sinx))=\lim_{x \rightarrow 0}[2(1+\tan^{2}2x)/(3+cosx)]=1/2\]
anonymous
  • anonymous
use expansion method it is the fastest.... tanx = x + x^3/3............sinx = x - x^3/3!
anonymous
  • anonymous
and from that the ans is 1/2 Math Physics and skywalker ur right..!!!!

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