Limits involving sin.

  • Thread starter mtayab1994
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  • #26
Dick
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Ok and 1-1 in the numerator gives 0 doesn't it?

NO. You get cos(u)-1. That's not zero.
 
  • #27
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so 2*cos(u)*sin(pi/6) is equal to cos(u)?
 
  • #28
Dick
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so 2*cos(u)*sin(pi/6) is equal to cos(u)?

Why are you asking that?
 
  • #29
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Why are you asking that?

because you said that 2*cos(u)*sin(pi/6) is the cos(u) part.
 
  • #30
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Ok i got the limit as 1/(2√3) is that correct?
 
  • #31
Dick
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because you said that 2*cos(u)*sin(pi/6) is the cos(u) part.

Are you saying that you don't see why it is equal to cos(u)???
 
  • #32
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Are you saying that you don't see why it is equal to cos(u)???

Is my answer of 1/(2√3) correct?
 
  • #33
Dick
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Is my answer of 1/(2√3) correct?

Yes, it is. That's the same thing l'Hopital would give you. You can always use l'Hopital to check these even if you can't use it to solve the problem.
 
  • #34
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Yes, it is. That's the same thing l'Hopital would give you. You can always use l'Hopital to check these even if you can't use it to solve the problem.

Yes thank you, I really appreciate your help :) .
 

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