Task:
Calculate lim 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)
\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
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