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

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 $$-1/x<\frac{-1}{1/M}=-M;\forall (-\delta). Since $$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?

• Yes the searched limit doesn't exist. – Dr. Sonnhard Graubner Dec 10 '18 at 20:16
• Can you use Maclaurin series? Logarithm expansion immediately gives you the resut – Alex Dec 10 '18 at 20:23
• @Alex I've never heard of the Maclaurin Series, but I'm curious how that proof would look like! Please share it. – Doesbaddel Dec 10 '18 at 20:25
• 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? – tommy1996q Dec 10 '18 at 20:30
• @tommy1996q It's just nomenclature. A Maclaurin series is just a Taylor expansion about $x=0$. – bob.sacamento Dec 10 '18 at 20:35

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

• Thank you very much, that shortens the proof significantly! – Doesbaddel Dec 11 '18 at 18:14
• @Doesbaddel You are welcome! That's strictly related to some general suggestions given here – user Dec 11 '18 at 18:15

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

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