Is this solution mathematically "legal"? I have the sequence
$$
a_n = \frac{n \cos n}{n^2 + 1}
$$
and I'm trying to evaluate the limit of $a_n$ as $n\to\infty$
$$
\begin{align*}
\lim_{n\to\infty}a_n&= \lim_{n\to\infty}\frac{n \cos n}{n^2 + 1}\\ &= \lim_{n\to\infty}\cos n \cdot \lim_{n\to\infty} \frac{n}{n^2 + 1}\\
\end{align*}
$$
Using L’hôpital’s rule twice on $\frac{n}{n^2+1}$
$$
\begin{align*}
\lim_{n\to\infty}a_n&= \lim_{n\to\infty}\cos n \cdot \lim_{n\to\infty} \frac{0}{2} \\ &= 0
\end{align*}
$$
Is there any flaw in this method?
 A: Your method has one minor technical error. In your method you are computing $\lim_{n \to \infty}{\cos(x)}$, which does not exist. The key to taking this limit is noting that $\cos(x)$ is bounded. That is,
\begin{align}
-1 \leq \cos(n) \leq 1
\end{align}
Hence,
\begin{align}
-\frac{n}{n^2+1} &\leq \frac{n\cos(n)}{n^2+1} \\
&\leq \frac{n}{n^2+1}.
\end{align}
Note that
\begin{align}
\lim_{n \to \infty}-\frac{n}{n^2+1} &= \lim_{n \to \infty}\frac{n}{n^2+1} \\
&= 0,
\end{align}
by L'Hopitals rule, for example. Hence, by the squeeze theorem,
\begin{align}
\lim_{n \to \infty}\frac{n\cos(n)}{n^2+1} &= 0.
\end{align}
A: It works if you expand your intention.
As others have mentioned, $\cos(x)$ doesn't have a limit as $x \to \infty$.
However any function has a set of limit points, and we can still give meaning to doing arithmetic with them. The set of limit points of $\cos(x)$ as $x \to \infty$ is the interval $[-1,1]$. And suitably interpreted, $[-1,1] \cdot 0 = 0$.
In short, there is something that is correct that resembles what you are trying to do. But what you literally said isn't right, and it's unclear if you had the right idea and didn't know how to express it or if you simply made an outright mistake.
A: This is not valid. To distribute limits, the individual limits must exist. As Matthew notes, the limit of $\cos n$ does not exist so you cannot distribute them in this way. However what you do know is that
$$\left|\frac{n\cos n}{n^2+1}\right| \le \frac{n}{n^2+1}$$
since $\cos$ is bounded by $\pm 1$. From here apply the squeeze theorem.
A: $$\frac{-n}{n^2+1} \leq a_n \leq \frac{n}{n^2+1}$$ You know what to do next.
A: No, it is not legal.
All you need to get the correct result
is $|\cos(n)| < 1$.
A: More simply, you can do that with asymptotic analysis and equivalents:


*

*$\cos n=O(1)$,

*$\dfrac n{n^2+1}\sim_\infty\dfrac1n$,


hence: $$\frac{n\cos n}{n^2+1}=O\Bigl(\frac1n\Bigr)\xrightarrow[n\to\infty]{}0.$$
