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

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). 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?

Answer

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