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

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

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

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

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