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?
Best Answer
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.