Computing lim sup and lim inf of $\exp(n\sin(\frac{n\pi}{2}))+\exp(\frac{1}{n}\cos(\frac{n\pi}{2}))$ and $\cosh(n\sin(\frac{n²+1}{n}\frac{\pi}{2}))$? It's the first time I encounter lim sup and lim inf and I only just know about their definitions. I have difficulties finding out about lim sup and lim inf of the following sequences
$$\exp\left(n\sin\left(\frac{n\pi}{2}\right)\right) + \exp\left(\frac{1}{n}\cos\left(\frac{n\pi}{2}\right)\right)$$
and 
$$\cosh\left(n\sin\left(\frac{n^2+1}{n}\frac{\pi}{2}\right)\right).$$
For the first one, I have the strong intuition that lim sup is infinite and maybe lim inf is 1? For the second one I'm really lost... Can you help me? Tell me how you proceed?
Thank you very much!!
 A: Note: After submitting the answer I realized that I've made a mistake. The conclusions that I draw are to be referred to $\sup$ and $\inf$ of the sequences, and not $\limsup$ and $\liminf$. I apologize for the error. I wanted to delete the answer, but then reconsidered it and I'll leave it here in case someone will find it useful. The reader should read those $\limsup$ and $\liminf$ as $\sup$ and $\inf$.
Let
$$a_n = \exp\left(n\sin\left(\frac{n\pi}{2}\right)\right) + \exp\left(\frac{1}{n}\cos\left(\frac{n\pi}{2}\right)\right)$$
We have that
$$\begin{align}
n = 2k &\implies \sin\left(n\frac\pi2\right) = 0\text{ and }\cos\left(n\frac\pi2\right) \in \{-1, 1\} \implies a_{2k} = 1 + \exp\left(\frac1n(-1)^k\right)\\
n = 2k + 1 &\implies \cos\left(n\frac\pi2\right) = 0\text{ and }\sin\left(n\frac\pi2\right) \in \{-1, 1\} \implies a_{2k+1} = \exp\left(n(-1)^k\right) + 1
\end{align}$$
We see that, as $k \to +\infty$, $a_{2k} \to 2$ and this subsequence attains its minimum and maximum values respectively when $k = 1$ and $k = 2$ (because it's oscillating around $2$ and the limit is $2$). So
$$\left\{a_{2k}\right\} \subseteq \left[1+\exp\left(-\frac12\right), 1 + \exp\left(\frac14\right)\right]$$
For the second one, we see that $n(-1)^k$ is divergent: as $k \to +\infty$, when $k$ is even $a_{2k+1} \to +\infty$, while when $k$ is odd we have that $a_{2k+1} \to -\infty$. Hence
$$\left\{a_{2k+1}\right\} \subseteq \left(1, +\infty\right)$$
Hence, we conclude that
$$\liminf a_n = 1,\qquad\limsup a_n = +\infty$$

For the second exercise, observe that $\cosh x \in [1, +\infty)$ and the minimum value is attained when $x = 0$. Also note that $\cosh x$ is an even function. We will now study the inner function to determine its behavior. Let
$$b_n = n\sin\left(\frac{n^2+1}n\frac\pi2\right)$$
As $n \to +\infty$ we can consider three subsequences of $b_n$:
$$\begin{align}
b_{n_k} &\to +\infty\\
b_{m_k} &\equiv 0\\
b_{l_k} &\to -\infty
\end{align}$$
We are not interested in finding $n_k$, $m_k$ and $l_k$, but it is fairly easy to prove that they exist. Hint: recall that $n \to +\infty$ and use the fact that $\sin x \in [-1, 1]$.
Putting it together:
$$\cosh\left(b_{n_k}\right) = \cosh\left(b_{l_k}\right) \to +\infty$$
$$\cosh\left(b_{m_k}\right) \equiv 1$$
Therefore, denoting the whole sequence with $a_n$, we have:
$$\liminf a_n = 1,\qquad\limsup a_n = +\infty$$
where $1$ is also a minimum.
