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?
 A: 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 
$$\lim_{x\to 0^+} \frac{\ln(1+2x)\sin x}{\sqrt {x^3}}=1\cdot2\cdot\lim_{x\to 0^+} \sqrt x.$$
A: 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 $x>0$,
$$
\frac{\ln(1+2x)\sin x}{\sqrt{x^3}}=2\cdot \frac{\ln(1+2x)}{2x}\cdot\frac{\sin x}{x}\cdot\frac{x^2}{\sqrt{x^3}}
$$
A: $$\frac{\ln(1+2x)\sin{x}}{\sqrt{x^3}}=\frac{\ln(1+2x)}{2x}\cdot\frac{\sin{x}}{x}\cdot2\sqrt{x}\rightarrow0.$$
A: If you want to, you should apply L'Hospital's rule twice:
$$\lim_{x\to 0^+} \frac{\ln(1+2x)\sin x}{\sqrt {x^3}}=\\
\lim_{x\to 0^+} \frac{\frac{2}{1+2x}\sin x+\ln(1+2x)\cos x}{1.5\sqrt {x}}=\\
\lim_{x\to 0^+} \frac{\frac{-4}{(1+2x)^2}\sin x+\frac{2}{1+2x}\cos x+\frac{2}{1+2x}\cos x-\ln(1+2x)\sin x}{\frac{0.75}{\sqrt {x}}}=0.
$$
