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It said as a hint to use the same proof but for sine to help, but I'm not really sure how to do that...
Last edited by Identity (2007-07-08 23:40:23)
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Are you allowed to use L'Hopital? That's the route I'd take.
Why did the vector cross the road?
It wanted to be normal.
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Actually I just figured out how to prove, I used
, divided through by tan, then took the limit of cos.But I would be happy if you could tell me what the L'Hopital is?
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Actually I just figured out how to prove, I used
, divided through by tan, then took the limit of cos.But I would be happy if you could tell me what the L'Hopital is?
That seems a bit complicated! Just write
L'Hopitals rule says that if f(x)-> 0, g(x)-> 0 as x-> a, then
That's much more "sophisticated" than is needed here.
I believe you are trying to use the squeeze theorem, no? I don't see how dividing through by tangent in that inequality will give you some form of tan(theta)/theta.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Hence, as
is squeezed in,Is that right?
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You are doing things without justification. How do you know that:
What happens when you divide by a negative in an inequality? What are you actually doing when you take a reciprocal? You kind of just gloss over these things.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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That was an inequality used in a previous proof (the one for sine) so I thought I might borrow it
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You can't just borrow and put things like that.
After borrowing you should manipulate it a little bit, fit it to your case, so nobody realize what had happened.
IPBLE: Increasing Performance By Lowering Expectations.
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