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Suppose $k_{n} \rightarrow k$ and $k$ is a constant. I want to show that $kk_{n} > \frac{k^2}{2}$ for a large enough $n$. Could someone give me feedback on my proof?

$\lim_{n\to\infty} kk_{n} = k \lim_{n\to\infty} k_{n} = k^2$, which is clearly bigger than $\frac{k^2}{2}$.

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    What if $k=0$? Otherwise, the idea is fine, but you ay want to tighten it: pick a specific $\epsilon$ and use the fact that $\lim\limits_{n\to\infty}kk_n = k^2$ to conclude, provided that $k\neq 0$, that there exists $N\gt 0$ such that for all $n\gt N$, $|kk_n - k^2|\lt \epsilon$ and use *that* to conclude $kk_n\gt k^2/2$.2012-03-06
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    Along the same lines: strictly speaking the result itself is false, for example if you choose $k_n = k = 0$2012-03-06
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    @ArturoMagidin: How about this? Let me take $\epsilon = \frac{k^2}{2}$. Then $|kk_{n} - k^2| < \epsilon \implies k - k^{2} / 2 = k^{2} / 2 < kk_{n}$2012-03-06
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    Provided you assume $k\neq 0$...2012-03-06

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