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?
 A: Hints: Your approaches are quite correct. Just explore them further.
For (a), try rewriting the interior expression as $\pi\cdot \dfrac{\sin n}{n} \cdot \cos n$.
For (b), divide top and bottom by $n$ and write it as $\dfrac{1}{\frac{\sin 2n}{n} - \frac{\cos \frac{n}{2} - 1}{n}}$
A: 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.
A: \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}
A: For (a), you can use a very similar approach.  Since $\cos(x)$ is continuous on $\mathbb{R}$, you can pull the limit inside, and you can rewrite the limit as:
\begin{align*}
\lim_{n\to 0} \cos\left(\frac{\pi}{n}\sin(n)\cos(n)\right) &= \lim_{n\to 0} \cos\left(\frac{\sin(n)}{n}\pi\cos(n) \right)\\
&= \cos\left(\lim_{n\to 0} \frac{\sin(n)}{n} \pi \cos(n) \right)
\end{align*}
For (b), try
\begin{align*}
\lim_{n\to 0} \frac{n}{\sin(2n) - \cos\left(\frac{n}{2}\right) +1}
&= \lim_{n\to 0} \frac{1}{\frac{\sin(2n)}{n} - \frac{\cos\left(\frac{n}{2}\right) -1}{n}}\\
&= \lim_{n\to 0} \frac{1}{2\frac{\sin(2n)}{2n} - \frac{1}{2}\frac{\cos\left(\frac{n}{2}\right) -1}{\frac{n}{2}}}
\end{align*}
(You will want to make sure you understand the steps in these equalities.  The first is dividing top and bottom by $n$, but the second is manipulating each term in the denominator to make them look like the limits you already know.)
And from here, your previous approach will prove useful.
