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anonymous

  • 5 years ago

How do I deal with this limit problem: limx->0 (tan6t/sin2t)? Converted tan to sin/cos, now what?

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  1. anonymous
    • 5 years ago
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    To evaluate limit, treat parts from numerator and denominator as two different functions, if you check that their limits as t approaches (note that you have error in your notation - you've written that \[x\] approaches zero) 0, you'll discover that it is zero. With this observation you can just use L'Hopital's rule to calculate limit.

  2. anonymous
    • 5 years ago
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    I get the parts dealing with sin(t)/t = 1 so I have about 2/3 of the answer. What do i do with the 1/cos6(t). The only rule I see is about cos(t)-1/t=0 (Yes, the x was a typo thanks.) According to my syllabus we are about four weeks away from the rule you are referring to so that doesn't help.

  3. anonymous
    • 5 years ago
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    Well, you can do it without L'Hopital's rule. To do so, you must (this step you've done as I see) rewrite tan as sine divided by cosine. Then, move bottom denominator (sin 2t) to fraction (it will be in fact cosecant). Now, you must cancel out this sin 2t because you can't divide by zero. In order to do this you must make fraction (you can just move numerator) of \[\sin 6t \over \sin 2t \]. This will lead you to solution.

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