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Task:

Calculate $$\lim_{x \to 0} \frac{\ln(1+2x)}{x^2}$$ with the help of l'Hospital's and Bernoullie's rule.

My thoughts:

Because $\mathcal{D}(f)=\{x\mid x\in\mathbb{R} \land x\neq 0\}$ the function is undefined for $0$ and therefore, I need to find out, whether the function has a limit or only one-sided limits. In order to do that, I'll just calculate the one sided limits. If $$\lim_{x \to 0^+} \frac{\ln(1+2x)}{x^2}\neq \lim_{x \to 0^-} \frac{\ln(1+2x)}{x^2} \implies \lim_{x \to 0} \frac{\ln(1+2x)}{x^2} \text{ doesn't exist}$$

$\lim_{x \to 0^-} \frac{\ln(1+2x)}{x^2}\overbrace{=}^{l'Hospital}=\lim_{x \to 0^-} \frac{2/(2x+1)}{2x}=\lim_{x \to 0^-}\frac{1}{x(2x+1)}\overbrace{=}^{product- rule}\underbrace{\lim_{x \to 0^-}\frac{1}{(2x+1)}}_{=1}\cdot \underbrace{\lim_{x \to 0^-}\frac{1}{x}}_{(*)}=1\cdot (*)=(*)$

$(*)$: If $x$ is small, than $1/x$ gets proportional bigger. Let $M>0$ and let $\delta = 1/M$. Than $-1/x<\frac{-1}{1/M}=-M;\forall (-\delta)<x<0$. Since $M$ can be arbitrarily large: $$\lim_{x \to 0^-} \frac1x=-\infty$$

$\lim_{x \to 0^-}$ analogue. $$\lim_{x \to 0^+} \frac{\ln(1+2x)}{x^2} = \cdots = \lim_{x \to 0^+} \frac1x=\infty$$ $\implies \lim_{x \to 0}$ doesn't exist.

Is this proof correct?

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    $\begingroup$ Yes the searched limit doesn't exist. $\endgroup$ – Dr. Sonnhard Graubner Dec 10 '18 at 20:16
  • $\begingroup$ Can you use Maclaurin series? Logarithm expansion immediately gives you the resut $\endgroup$ – Alex Dec 10 '18 at 20:23
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    $\begingroup$ @Alex I've never heard of the Maclaurin Series, but I'm curious how that proof would look like! Please share it. $\endgroup$ – Doesbaddel Dec 10 '18 at 20:25
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    $\begingroup$ Maybe you have heard of Taylor series? It’s practically the same thing. Actually I don’t know why they are called differently, anybody knows? $\endgroup$ – tommy1996q Dec 10 '18 at 20:30
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    $\begingroup$ @tommy1996q It's just nomenclature. A Maclaurin series is just a Taylor expansion about $x=0$. $\endgroup$ – bob.sacamento Dec 10 '18 at 20:35
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Yes your evaluation is fine, to check it by standard limits, we have that

$$ \frac{\ln(1+2x)}{x^2}=\frac{\ln(1+2x)}{2x}\frac{2}{x}\to 1\cdot\pm\infty$$

therefore the limit doesn't exist.

For the proof of the standard limit refer to

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  • $\begingroup$ Thank you very much, that shortens the proof significantly! $\endgroup$ – Doesbaddel Dec 11 '18 at 18:14
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    $\begingroup$ @Doesbaddel You are welcome! That's strictly related to some general suggestions given here $\endgroup$ – user Dec 11 '18 at 18:15
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Since as $x \to 0 \ \log (1 + 2x) \to 0$, you can expand $\log$ function around $x=0$ to get (first term is enough) $\log (1+2x) \sim 2x$, and the fraction becomes $\frac{2}{x}$ that certainly diverges

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  • $\begingroup$ Thanks, way shorter than my thoughts. $\endgroup$ – Doesbaddel Dec 11 '18 at 18:13

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