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Re: Math Question #1 [message #357822 is a reply to message #357810] Sun, 09 November 2008 09:26 Go to previous messageGo to previous message
StealthEye is currently offline  StealthEye
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Registered: May 2006
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Your answer is correct, carrier, however the method is not. Try the same with assuming the divident is "some number" and your approach will lead you to the limit being 0, which is not correct.

I can't really come up with a correct prove either however. Closest I can get is to say that y=sin(a) behaves like y=a for x~0 and b=cos(a) behaves like 1 in that interval. Computing the limit after substituting those gives lim=+inf. This, however, is not solid prove either (actually, it's just disguised 'l hopital).

I would expect it would be possible to rewrite the 1-cos(x) to some sin variant or vice versa and then solve it to get rid of (one of) the trig functions. 'l hopital is much easier. Sad

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