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@Brian Charlesworth
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Another way is to put πx=y and by first principles we can say the limit equals negative of the derivative of tany at y=3π. This is similar to L'Hospital but not completely.
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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2^{34}
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We could just use L'Hopital's rule, but I'm assuming that you are looking for an approach that does not use this rule.
Let u=x−31. Then u→0 as x→31 and x=u+31, in which case
3−tan(πx)=3−tan(πu+3π)=3−1−tan(πu)tan(3π)tan(πu)+tan(3π)=
3−1−3tan(πu)tan(πu)+3=1−3tan(πu)3−3tan(πu)−tan(πu)−3=1−3tan(πu)−4tan(πu).
Thus the original limit can be written as
limu→03u∗(1−3tan(πu))−4tan(πu)=
limu→03∗(πu)−4πtan(πu)∗limu→01−3tan(πu)1=−34π∗limw→0wtan(w)∗1=−34π,
where w=πu→0 as u→0. Note also that (1−3tan(πu))→1 as u→0.
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Nice! Approaches without using L'hopital's rule are always really good.
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Thanks! L'Hopital's is really useful if you want a quick result, but I like the challenge of trying to find an alternative approach. :)
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πx=y and by first principles we can say the limit equals negative of the derivative of tany at y=3π. This is similar to L'Hospital but not completely.
Another way is to put3−4π
-4π/3