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I need to find this limit:

$\displaystyle\lim_{v\to180}\frac{360\cos\left(\dfrac{v}{2}\right)}{180-v}$, with $v$ in degrees.

I have tried to do this:

$\displaystyle\lim_{v\to180}\frac{360\cos\left(\dfrac{v}{2}\right)}{180-v}=^{L'H}\displaystyle\lim_{v\to180}\frac{-\frac{360}{2}\sin\left(\dfrac{v}{2}\right)}{-1}=180\sin\left(\dfrac{v}{2}\right)=180$.

This is wrong though, because its limit is actually $\pi$. I realise that $\pi$ radians = 180 degrees, but as far as i can tell nothing is linking degrees and radians in the formula. Could someone show me how to find this limit?

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    $\begingroup$ And now you've found that differential/integral calculus just does not work "as supposed" when instead radians we use degrees...! In fact, if you're going to work with degrees, you will have to use the equivalence $$180\;deg. =\pi\;rad.$$otherwise, for example, $\;\cos\frac{180}2=\cos90\neq 0\;$ since $\;\cos 90\;radians\neq 0\;$ ...! $\endgroup$
    – DonAntonio
    Nov 12, 2013 at 16:05
  • $\begingroup$ @DonAntonio Ah thank you! Is this because I can't take the derivative of a trig function if it is in degrees, hence L'Hôpital's rule doesn't work? $\endgroup$ Nov 12, 2013 at 16:09
  • $\begingroup$ Yes...or, in fact, because $\;\cos(x\;rad.)=\cos\frac{180x}{\pi}\;deg.\;$. Now use the chain rule....but, of course, it is too cumbersome. $\endgroup$
    – DonAntonio
    Nov 12, 2013 at 16:11
  • $\begingroup$ The best way to think of it is as follows: the functions $\sin$, $\cos$, $\tan$, etc. are functions that take numbers and give numbers. Those numbers they take can be seen as representing angle measure in radians only. If you want to deal with degrees, you always need a conversion factor: $\cos x^\circ = \cos \left(\frac \pi {180} x\right)$. To put a finer point on it, $\cos 90$ simply does not ever mean the cosine of a right angle (except perhaps in a sloppy geometry class). It's the cosine of a $90$-radian angle. $\endgroup$
    – dfeuer
    Nov 12, 2013 at 17:34

2 Answers 2

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Translate to radians in order to make sense of the usual differential stuff we know:

$$\lim_{x\to \pi}\frac{2\pi\cos\frac x2}{\pi -x}\stackrel{\text{l'Hôpital}}=\lim_{x\to\pi}\frac{-\pi\sin\frac x2}{-1}=\pi$$

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  • $\begingroup$ Thank you. Would we have a way to do this if we didn't know about radians? By the way you might want to change l'Hospital to l'Hopital. $\endgroup$ Nov 12, 2013 at 16:14
  • $\begingroup$ Both forms, with and without "s", are correct...and without knowing about radians we couldn't possibly do sense with the differentials... $\endgroup$
    – DonAntonio
    Nov 12, 2013 at 16:18
  • $\begingroup$ Oh sorry, I didn't know that, but now I do :). What about ways of doing it without differentials? $\endgroup$ Nov 12, 2013 at 16:20
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$$cos(\frac{v}{2})=sin\frac{1}{2}(180-v)$$let $t=180-v$ then limit changes to $$\lim_{t\to0} \frac{360sin(\frac{t}{2})}{t}$$multiply and divide by $2$ to get the limit as $180$

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