limits of sequences exponential and factorial: $a_n=e^{5\cos((\pi/6)^n)}$ and $a_n=\frac{n!}{n^n}$ Compute the limit $\lim\limits_{n\to\infty}a_n$ for the following sequences:
(a) $a_n=e^{5\cos((\pi/6)^n)}$
(b) $a_n=\frac{n!}{n^n}$
For part (a) do I just take the limit of the exponent part and then the answer would be $e$ raised to whatever the limit is?
And would the limit be $1$ or $-1$? because $\cos$ goes between those two.
For part $b$ it is in the form of infinity over infinity but how do you take the derivative of $n!$? Will it ever break out of infinity over infinity?
 A: (a) 
$$
\lim_{n \to \infty} e^{5 \cos((\frac{\pi}6))^n} = e^{\lim_{n \to \infty} 5 \cos((\frac{\pi}6))^n} = e^{5 \cos(\lim_{n \to \infty}(\frac{\pi}6)^n)} = e^{5 \cdot 1} = e^5
$$
(b)
$$
 \frac{n!}{n^n} = \frac{1\cdot 2 \cdots n}{n\cdot n \cdot n} \leq \frac{1}{n}
$$
From here 
$$
\lim_{n \to \infty}  \frac{n!}{n^n} = 0
$$
since 
$$
\lim_{n \to \infty} \frac{1}{n} = 0
$$
A: Note that 
$$\lim_{n\to\infty}\left(\frac{\pi}{6}\right)^n=0.$$ 
Since $\cos x$ is continuous, it follows that 
$$\lim_{n\to\infty}\cos\left(\left(\frac{\pi}{6}\right)^n\right)=\cos(0)=1.$$ 
Since the function $e^x$ is continuous, it follows that the required limit is $e^5$.
For the second problem, please see the answer by UrošSlovenija.
A: For the second one
(I've done this before btw),
$n! = \prod_{i=1}^n i
=\prod_{i=1}^n (n+1-i)
$
so
$\begin{align}
n!^2 
&= \prod_{i=1}^n i(n+1-i)\\
&= \prod_{i=1}^n (i(n+1)-i^2)\\
&= \prod_{i=1}^n ((n+1)/2)^2-(n+1)/2)^2+ i(n+1)-i^2)\\
&= \prod_{i=1}^n ((n+1)/2)^2-((n+1)/2-i)^2)\\
&\le \prod_{i=1}^n ((n+1)/2)^2)\\
&= ((n+1)/2)^{2n}
\end{align}
$
so $n! < (\frac{n+1}{2})^{n}
$
and
$\frac{n!}{n^n}
\le (\frac{n+1}{2n})^{n}
$
and this $\to 0$ as $n \to \infty$.
