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One of my students approached me with a question about a limit. (She doesn't know the idea of a limit.) I was able to find the limit quite easily by using L'Hopital's Rule. However, they don't know about L'Hopital's Rule, and so a proof cannot use it. The limit is as follows:

$$\lim_{x \to \infty} \frac{\ln^2x + 2\ln x}{x}=0$$

I must be missing a simple trick. The proof cannot use L'Hopital's Rule, nor can it be a formal $\varepsilon\delta$-proof. Neither of these ideas have been introduced in the course. We have done Taylor Series.

Can anyone see a simple way, of finding this limit?

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Let $x=e^u$. We are looking at $\frac{u^2+2u}{e^u}$.

For the $\frac{u^2}{e^u}$ part, use the fact that for positive $u$, we have $e^u\gt \frac{u^3}{3!}$. (This comes from the Taylor series.)

The $\frac{2u}{e^u}$ part is done in the same way.

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  • $\begingroup$ That's nice André. Thank you very much. $\endgroup$ – Fly by Night Feb 3 '14 at 20:33
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Feb 3 '14 at 20:40
  • $\begingroup$ To keep this at the U.S. high school level (so no Taylor series, or even derivatives), for the $\frac{u^2}{e^u}$ and $\frac{2u}{e^u}$ parts make use of the fact that $e^u > 2^u = \left( 2^{10} \right)^{(u/10)} > \left( 10^3\right)^{(u/10)} = 10^{\frac{3u}{10}}$ and then look at what happens when $u = 10, \; 100, \; 1000, \; 10000, \; \ldots$ $\endgroup$ – Dave L. Renfro Feb 3 '14 at 20:41

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