How to find $\lim_{n\rightarrow 0} \cos(\frac{\pi}{n} \sin n \cos n)$ and $\lim_{n\rightarrow 0} \frac{n}{\sin(2n) - \cos(\frac{n}{2}) +1}$ 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?

\begin{align}
(a)\quad\lim_{n\rightarrow 0} \cos\left(\frac{\pi\sin n \cos n}{n}\right)&=\cos\left(\lim_{n\rightarrow 0}\frac{\pi\sin n \cos n}{n}\right)\\
&=\cos\left(\lim_{n\rightarrow 0}\frac{\pi\sin n \cos n}{n}\right)\\
&=\cos\left(\pi\lim_{n\rightarrow 0}\cos n\frac{\sin n}{n}\right)\\
&=\cos(\pi)\\
&=-1.
\end{align}
Note that, it is valid to move the limit since cosine function is continuous.
\begin{align}
(b)\quad\lim_{n\rightarrow 0}  \dfrac{\dfrac nn}{\dfrac{2\sin2n}{2n} -\left(\dfrac{ \cos\frac{n}{2}-1}{2\cdot\frac{n}{2}}\right)}=\frac{1}{2}.
\end{align}
