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August 7th, 2012, 01:21 AM   #1
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Work check on a trig. limit

I am hoping someone can double check to make sure I didn't do anything illegal (mathematically), especially in step 2 and 3. I am not sure if I just made things up, or if I am allowed to work it this way by some math principles?

Thanks!
Robert

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August 7th, 2012, 01:27 AM   #2
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Re: Work check on a trig. limit

You cannot factor the 3 out of the sine function's argument (even though this coincidentally produces the correct result). I would write:

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August 7th, 2012, 01:43 AM   #3
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Re: Work check on a trig. limit

Do you know why we can not factor out the 3 if it produces the same answer? I have done three problems like this and have got the answer correct. Check this one out:



I am wondering why its wrong when I am getting them right?
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August 7th, 2012, 01:53 AM   #4
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Re: Work check on a trig. limit

The reason I would give is that for , we have:



The second problem, I would solve as follows (barring the use of L'H˘pital's rule):

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August 7th, 2012, 01:14 PM   #5
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Re: Work check on a trig. limit

(x -> 0) Lim sin(kx)/x = klim sin(kx)/kx = k. That is why it works.
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