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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. $\endgroup$ – Matthew Leingang Aug 18 '15 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. $\endgroup$ – Cameron Williams Aug 18 '15 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. $\endgroup$ – MartianInvader Aug 18 '15 at 18:16
  • $\begingroup$ Completely missed that too. Thanks. $\endgroup$ – rgarci0959 Aug 18 '15 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 '15 at 3:48
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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.

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

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

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

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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}

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    $\begingroup$ Good solution. I wouldn't call the error minor, though. It belies a misconception about what nonexistent limits represent. $\endgroup$ – Matthew Leingang Aug 18 '15 at 16:56
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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.

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  • $\begingroup$ 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. $\endgroup$ – Eric Wofsey Aug 18 '15 at 19:03
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$$\frac{-n}{n^2+1} \leq a_n \leq \frac{n}{n^2+1}$$ You know what to do next.

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No, it is not legal.

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

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