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Is it true that $$\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = c_1 \implies \lim_{x \rightarrow \infty} \frac{g(x)}{f(x)} = c_2$$ where $0 < c_1,c_2 < \infty$?

If yes, how can we prove that?

Edit: $f(x)$ and $g(x)$ are non-decreasing.

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  • $\begingroup$ Yes it is true. In fact $c_{2}=1/c_{1}$. Use algebra of limits on identity $g(x)/f(x)=1/(f(x)/g(x))$. $\endgroup$ Mar 25 '16 at 7:38
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Hint. One may use the fact that $x \mapsto \dfrac1x$ is continuous over $(0,\infty)$.

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  • $\begingroup$ Can you give a complete answer? My calculus is really rusty. $\endgroup$ Mar 23 '16 at 18:56
  • $\begingroup$ One may recall that if $u$ is continuous over a certain set then $\displaystyle \lim_{\infty}v(x)=\ell$ implies $\displaystyle\lim_{\infty}u(v(x))=u(\ell)$. Here you can take $u(x)=1/x$ and $v(x)=f(x)/g(x)$ giving $c_2=1/c_1$. $\endgroup$ Mar 23 '16 at 21:05

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