# How to find the limit of <munder> <mo movablelimits="true" form="prefix">lim <mrow cla

How to find the limit of $\underset{n\to \mathrm{\infty }}{lim}\left(n\mathrm{sin}\frac{n\pi }{2}\mathrm{cos}\frac{1}{n}\right)$
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Punktatsp
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 $\mathrm{cos}\left(1/n\right)\to 1$; moreover, since $\mathrm{cos}\left(1/n\right)=\mathrm{cos}\left(1\right)>\mathrm{cos}\left(\pi /2\right)=0$ we have that $\mathrm{cos}\left(1/n\right)>0$ for every $n\ge 1$. Now note that
$\mathrm{sin}\left(n\pi /2\right)=\left\{\begin{array}{ll}±1& n\equiv 1,3\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}4\\ 0& n\equiv 0,2\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}4\end{array}$
so if we take $n\to \mathrm{\infty }$ with n=2m then
$2m\mathrm{sin}\left(\frac{2m\pi }{2}\right)\mathrm{cos}\left(\frac{1}{2m}\right)=2m\mathrm{cos}\left(\frac{1}{2m}\right)\mathrm{sin}\left(m\pi \right)=0$
while if we take $n\to \mathrm{\infty }$ with n odd, then
$|n\mathrm{sin}\left(\frac{n\pi }{2}\right)\mathrm{cos}\left(\frac{1}{n}\right)|=|\left(2k+1\right)\mathrm{sin}\left(\frac{\left(2k+1\right)\pi }{2}\right)\mathrm{cos}\left(\frac{1}{2k+1}\right)|=|\left(2k+1\right)\mathrm{cos}\left(\frac{1}{2k+1}\right)|$
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