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While I tried to show, that any differentiable function is also steady, I found this lemma but wasn't able to show it. Is it true?

Given two sequences $a_n$ and $b_n$, such that $\lim_{n\to\infty}b_n=0$ and $\lim_{n\to\infty}\frac{a_n}{b_n}$ exist, then $\lim_{n\to\infty}a_n=0.$

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    $\begingroup$ Have you already got a theorem to the effect that if $\lim x_n$ is the real number $x$, and $\lim y_n$ is the real number $y$, then $\lim x_ny_n =xy$? $\endgroup$ Nov 11, 2011 at 8:21
  • $\begingroup$ @André: yes. We already proved that. $\endgroup$
    – FUZxxl
    Nov 11, 2011 at 9:10
  • $\begingroup$ Then you can simply let $x_n=b_n$ and $y_n=\frac{a_n}{b_n}$. Note that $x_ny_n=a_n$ and the desired result falls out. $\endgroup$ Nov 11, 2011 at 12:36

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Hint: Show the stronger result (but with a simpler proof) that if $b_n\to0$ and $(c_n)$ is bounded then $b_nc_n\to0$. Then apply it to $c_n=a_n/b_n$.

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What do you mean by exist? I guess you allow only finite real numbers as limits (otherwise the lemma is trivially false).

You can prove it as follows: suppose that $\limsup a_n=c>0$, hence there is a subsequence, which I denote again by $a_n$ such that $\lim a_n=c$. Along this subsequence you get $\lim\frac{a_n}{b_n}=\infty$ (to be really formal you should take also a suitable subsequence of $b_n$)... this is a contradiction. Analogously, you get a contradiction, assuming that $\liminf a_n=c<0$.

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