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?

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    $\begingroup$ $\lim_{n\to\infty} \cos(n)$ does not exist. $\endgroup$ Aug 18, 2015 at 16:38
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    $\begingroup$ To expand upon @MatthewLeingang's comment: you can only distribute limits when the individual pieces have limits. In this case, you should apply the squeeze theorem. $\endgroup$ Aug 18, 2015 at 16:39
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    $\begingroup$ Also, you can't use L'Hopital's rule twice on $\frac{n}{n^2+1}$, only once. After that, it's not indeterminate. $\endgroup$ Aug 18, 2015 at 18:16
  • $\begingroup$ Completely missed that too. Thanks. $\endgroup$
    – rgarci0959
    Aug 18, 2015 at 18:17
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    $\begingroup$ In my opinion using LHR for sequences is a very very bad idea. You miss out on the interesting theorems which have been devised for evaluation of limits of sequences. If you really wish to use LHR go for the sequence version called Stolz theorem. $\endgroup$
    – Paramanand Singh
    Aug 19, 2015 at 3:48

6 Answers 6


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.


More simply, you can do that with asymptotic analysis and equivalents:

  • $\cos n=O(1)$,
  • $\dfrac n{n^2+1}\sim_\infty\dfrac1n$,

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


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}


\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}

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    $\begingroup$ Good solution. I wouldn't call the error minor, though. It belies a misconception about what nonexistent limits represent. $\endgroup$ Aug 18, 2015 at 16:56

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.

  • $\begingroup$ Note that the assertion that sets of limit points are preserved by continuous functions (in this case, multiplication) is somewhat subtle and makes essential use of compactness. $\endgroup$ Aug 18, 2015 at 19:03

$$\frac{-n}{n^2+1} \leq a_n \leq \frac{n}{n^2+1}$$ You know what to do next.


No, it is not legal.

All you need to get the correct result is $|\cos(n)| < 1$.


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