I have to find the limit $$\lim_{x\to 0^+} \frac{\ln(1+2x)\sin x}{\sqrt {x^3}} $$

Now, I tried using L'Hôpital's rule, but it doesn't lead anywhere.

Manually trying to convert the functions to another form doesn't seem to go anywhere either, so I determined that the limit must be undefined.

However, I cannot prove it. What can I do?

  • 2
    $\begingroup$ Using Taylor, the numerator is $2x\cdot x+o(x^3)$ and the limit is $0$. $\endgroup$ – Yves Daoust Jan 30 at 16:37
  • $\begingroup$ We haven't been taught about Taylor yet, I should add this to my question. However, thanks for the answer, i'll research Taylor on my own. $\endgroup$ – sdds Jan 30 at 16:40
  • $\begingroup$ To answer the question in the title, see for example here $\endgroup$ – BlueRaja - Danny Pflughoeft Jan 30 at 21:37
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Use the facts that (1) $\lim_{x\to 0} \frac{\ln(1+2x)}{2x} = 1$, (2) $\lim_{x\to 0} \frac{\sin x}{x} = 1$, and (3) $\lim_{x\to 0^+} \frac{x^2}{\sqrt{x^3}} = 0$ to show the limit is $2\cdot 1\cdot 0 = 0$.

To do so, rewrite, for $x>0$, $$ \frac{\ln(1+2x)\sin x}{\sqrt{x^3}}=2\cdot \frac{\ln(1+2x)}{2x}\cdot\frac{\sin x}{x}\cdot\frac{x^2}{\sqrt{x^3}} $$

  • $\begingroup$ Thank you, this conversion never occurred to me $\endgroup$ – sdds Jan 30 at 16:37
  • 1
    $\begingroup$ @sdds You're welcome! $\endgroup$ – Clement C. Jan 30 at 16:40



It is known that

$$\lim_{x\to 0^+}\frac{\sin x}x=1$$ and $$\lim_{x\to 0^+}\frac{\log(1+x)}x=1=\lim_{x\to 0^+}\frac{\log(1+2x)}{2x}.$$


$$\lim_{x\to 0^+} \frac{\ln(1+2x)\sin x}{\sqrt {x^3}}=1\cdot2\cdot\lim_{x\to 0^+} \sqrt x.$$

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    $\begingroup$ It's true, but how is that different from the current two answers? $\endgroup$ – Clement C. Jan 30 at 16:45

If you want to, you should apply L'Hospital's rule twice: $$\lim_{x\to 0^+} \frac{\ln(1+2x)\sin x}{\sqrt {x^3}}=\\ \lim_{x\to 0^+} \frac{\frac{2}{1+2x}\sin x+\ln(1+2x)\cos x}{1.5\sqrt {x}}=\\ \lim_{x\to 0^+} \frac{\frac{-4}{(1+2x)^2}\sin x+\frac{2}{1+2x}\cos x+\frac{2}{1+2x}\cos x-\ln(1+2x)\sin x}{\frac{0.75}{\sqrt {x}}}=0. $$


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