Finding the limit of $\lim_{v\to180}\frac{360\cos\left(\frac{v}{2}\right)}{180-v}$

Question

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?

Answer 1

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

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

Answer 2

$$cos(\frac{v}{2})=sin\frac{1}{2}(180-v)$$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 $180$