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