TITLE: How to find $\lim_{n\rightarrow 0} \cos(\frac{\pi}{n} \sin n \cos n)$ and $\lim_{n\rightarrow 0} \frac{n}{\sin(2n) - \cos(\frac{n}{2}) +1}$
QUESTION [5 upvotes]: How can I determine these limits:
$$ a) \lim_{n\rightarrow 0} \cos(\frac{\pi}{n} \sin n \cos n)$$
$$ b) \lim_{n\rightarrow 0}  \frac{n}{\sin(2n) - \cos(\frac{n}{2}) +1}$$
Note I cannot use l'Hospital. But I know that $\lim_{n\rightarrow 0} \frac{\sin n}{n} = 1$ and $\lim_{n\rightarrow 0} \frac{\cos n -1}{n} = 0$
I already found the limit of $\lim_{n\rightarrow 0}  \frac{\sin(3n) \sin(2n)}{n^2}$:
$$\lim_{n\rightarrow 0}  \frac{\sin(3n) \sin(2n)}{n^2} 
= \lim_{n\rightarrow 0}  \frac{\sin(3n)}{n}  \lim_{n\rightarrow 0} \frac{ \sin(2n)}{n} = 
3 \lim_{n\rightarrow 0}  \frac{\sin(3n)}{3n}  2\lim_{n\rightarrow 0} \frac{ \sin(2n)}{2n} = 3 \cdot 2 = 6$$
Can I use a similar approach here?

REPLY [4 votes]: Hint For part (a) you could also use the double angle formula in reverse 
$$ 
\begin{align}
\frac{\pi}{n} \sin n \cos n = \pi\frac{\sin 2n}{2n}
\end{align}
$$
Which you already know the limit to.
Hint Part (b) can be done by dividing the numerator and denominator by n.