# How can I claim a one-sided limit doesn't exist?

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?

• Using Taylor, the numerator is $2x\cdot x+o(x^3)$ and the limit is $0$. – Yves Daoust Jan 30 at 16:37
• 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. – sdds Jan 30 at 16:40
• To answer the question in the title, see for example here – BlueRaja - Danny Pflughoeft Jan 30 at 21:37
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Hint:

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

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

$$\frac{\ln(1+2x)\sin{x}}{\sqrt{x^3}}=\frac{\ln(1+2x)}{2x}\cdot\frac{\sin{x}}{x}\cdot2\sqrt{x}\rightarrow0.$$

It is known that

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

Then

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

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

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