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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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    Provided you assume $k\neq 0$...2012-03-06

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It's better to prove a general statement that can be applied to many such problems.

Fact: if $\lim_{n\to\infty} x_n=L$ and $a, then $x_n>a$ for all sufficiently large $n$.

Proof sketch: use the definition of limit with $\epsilon=L-a$.

Application to the specific problem. We have $\lim_{n\to\infty} (kk_n)=k^2>k^2/2$ provided that $k\ne 0$. It follows that in this case $kk_n>k^2/2$ for all sufficiently large $n$.