Limit of an oscillating sequence Let $a$ be any real number. Then, how do I compute the 
$$
\limsup_n \left|\cos\big(a\sqrt{n^2+1}\big) \right|^{1/n}\:?
$$
Also, how do I compute the value when $a$ is any complex value? The sequence is harshly oscillating... so I cannot find a way to approach

I'll find the limit in the most general case when $a$ is complex.
Let $a=x+iy$ and the $n^{th}$ term of the sequence be denoted by $s_n$
\begin{align*}
s_n &= \left| \cos \left( a \sqrt{n^2+1} \right) \right|^{1/n} \\
&= \left|  \frac{ e^{ia\sqrt{n^2+1}} + e^{-ia\sqrt{n^2+1}} }{2} \right|^{1/n} \\
&=\left|e^{ia\sqrt{n^2+1}} \right|^{1/n} \, \left| \frac{1+e^{-2ia\sqrt{n^2+1}}}{2} \right|^{1/n} \\
&= e^{ \frac{-y \sqrt{n^2+1}}{n}} \, \left| \frac{1+e^{-2ia\sqrt{n^2+1}}}{2} \right|^{1/n} \\
\end{align*}
Let the second term (without the power) be denoted by  $t_n$ .
\begin{align*}
t_n &= \left| \frac{1+e^{2y\sqrt{n^2+1}} e^{-2ix\sqrt{n^2+1}} }{2} \right|\\
&= \frac{1}{2} \left( 1+ e^{4y\sqrt{n^2+1}}+2e^{2y\sqrt{n^2+1}}\cos(2x\sqrt{n^2+1}) \right)^{1/2}
\end{align*}
Now observe that 
\begin{equation*}
u_n\leq t_n \leq v_n
\end{equation*}
where
\begin{align*}
u_n &= \left| \frac{1-e^{2y\sqrt{n^2+1}}}{2} \right| \\
v_n &= \left| \frac{1+e^{2y\sqrt{n^2+1}}}{2} \right|
\end{align*}
so we get for our original sequence $s_n$
\begin{equation*}
 \left( e^{-y \sqrt{n^2+1}}u_n \right) ^{1/n} \leq s_n \leq \left( e^{-y \sqrt{n^2+1}}v_n \right) ^{1/n}
\end{equation*}
which gives
\begin{equation*}
\sinh \left( y\sqrt{n^2+1} \right) ^{1/n} \leq s_n \leq \cosh \left( y\sqrt{n^2+1} \right) ^{1/n}
\end{equation*}
Now in the limit $n\to \infty$, both of the sequences that sandwich our original sequence tend to $exp(y)$. Hence we get our answer
\begin{equation*}
 \lim_{n\to \infty} s_n = e^y
\end{equation*}
