I was wondering how this worked when submitted, anything over 0 was infinite.
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Have you tried l'Hopital's rule? – ajotatxe Apr 30 '14 at 05:20
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Are you familiar with indeterminate forms? Sure, if your limit looks like $\frac{k}{0}$ and $k \ne 0$ then you're going to have an infinite limit, but if the functions on the top and bottom both tend to zero then you need to do more work (e.g. applying L'Hôpital's rule before you can determine the limit. And sure enough, $\lim_{x \to 0} \cos (\frac{\pi}{2} - x) = 0$. – Clive Newstead Apr 30 '14 at 05:22
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1$$\cos{(\frac{\pi}{2}-x)}=\sin x$$ and see this – evil999man Apr 30 '14 at 05:22
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Why was this answer converted to comment? – evil999man Apr 30 '14 at 05:23
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the answer is 1 though! – chrissy kwon Apr 30 '14 at 05:29
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I'm not sure because the name of the worksheet I was doing was Limits at Removable Discontinuities using trig, from AB calc – chrissy kwon Apr 30 '14 at 05:30