How to find the limit of $\underset{n\to \mathrm{\infty}}{lim}(n\mathrm{sin}\frac{n\pi}{2}\mathrm{cos}\frac{1}{n})$

Sonia Gay
2022-06-22
Answered

How to find the limit of $\underset{n\to \mathrm{\infty}}{lim}(n\mathrm{sin}\frac{n\pi}{2}\mathrm{cos}\frac{1}{n})$

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Punktatsp

Answered 2022-06-23
Author has **22** answers

The basic strategy for showing that limits of this kind do not exist is to approach infinity along a different path; usually we consider what happens as n approaches $\mathrm{\infty}$ from the evens and the odds. I've provided an example of how to do this with your problem as provided below.

First note that as $n\to \mathrm{\infty}$, $1/n\to 0$ so $\mathrm{cos}(1/n)\to 1$; moreover, since $\mathrm{cos}(1/n)=\mathrm{cos}(1)>\mathrm{cos}(\pi /2)=0$ we have that $\mathrm{cos}(1/n)>0$ for every $n\ge 1$. Now note that

$\mathrm{sin}(n\pi /2)=\{\begin{array}{ll}\pm 1& n\equiv 1,3\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}4\\ 0& n\equiv 0,2\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}4\end{array}$

so if we take $n\to \mathrm{\infty}$ with n=2m then

$2m\mathrm{sin}\left(\frac{2m\pi}{2}\right)\mathrm{cos}\left(\frac{1}{2m}\right)=2m\mathrm{cos}\left(\frac{1}{2m}\right)\mathrm{sin}(m\pi )=0$

while if we take $n\to \mathrm{\infty}$ with n odd, then

$|n\mathrm{sin}\left(\frac{n\pi}{2}\right)\mathrm{cos}\left(\frac{1}{n}\right)|=|(2k+1)\mathrm{sin}\left(\frac{(2k+1)\pi}{2}\right)\mathrm{cos}\left(\frac{1}{2k+1}\right)|=|(2k+1)\mathrm{cos}\left(\frac{1}{2k+1}\right)|$

which evidently approaches $\mathrm{\infty}$ as $n\to \mathrm{\infty}$. Since we have approached $\mathrm{\infty}$ along two different paths (namely n even and n odd) that produce different values, we conclude that the limit does not exist

First note that as $n\to \mathrm{\infty}$, $1/n\to 0$ so $\mathrm{cos}(1/n)\to 1$; moreover, since $\mathrm{cos}(1/n)=\mathrm{cos}(1)>\mathrm{cos}(\pi /2)=0$ we have that $\mathrm{cos}(1/n)>0$ for every $n\ge 1$. Now note that

$\mathrm{sin}(n\pi /2)=\{\begin{array}{ll}\pm 1& n\equiv 1,3\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}4\\ 0& n\equiv 0,2\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}4\end{array}$

so if we take $n\to \mathrm{\infty}$ with n=2m then

$2m\mathrm{sin}\left(\frac{2m\pi}{2}\right)\mathrm{cos}\left(\frac{1}{2m}\right)=2m\mathrm{cos}\left(\frac{1}{2m}\right)\mathrm{sin}(m\pi )=0$

while if we take $n\to \mathrm{\infty}$ with n odd, then

$|n\mathrm{sin}\left(\frac{n\pi}{2}\right)\mathrm{cos}\left(\frac{1}{n}\right)|=|(2k+1)\mathrm{sin}\left(\frac{(2k+1)\pi}{2}\right)\mathrm{cos}\left(\frac{1}{2k+1}\right)|=|(2k+1)\mathrm{cos}\left(\frac{1}{2k+1}\right)|$

which evidently approaches $\mathrm{\infty}$ as $n\to \mathrm{\infty}$. Since we have approached $\mathrm{\infty}$ along two different paths (namely n even and n odd) that produce different values, we conclude that the limit does not exist

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