How to solve this limit involving sine and log? I've tried L'Hopital's Rule but the differentiated numerator involves cos(1/x) which does not exist when x approaches 0.
$$ \lim_{x\to 0^+} \frac{x^2sin\frac{1}{x}}{\ln(1+2x)}$$

Note that, by applying the Squeeze theorem you get
\begin{align}
&-1 \le \sin x \le 1\\
&\Rightarrow \lim_{x \to \infty} \frac{-1}{x} \le \lim_{x \to \infty} \frac{\sin x}{x} \le \lim_{x \to \infty} \frac{1}{x}\\
&\Rightarrow 0 \le \lim_{x \to \infty} \frac{\sin x}{x} \le 0\\
&\Rightarrow \lim_{x \to \infty} \frac{\sin x}{x} = 0\\
&\Rightarrow \lim_{y \to 0+} y \sin \frac{1}{y} = 0
\end{align}
so
\begin{align}
\lim_{x\to 0^+} \frac{x^2\sin\frac{1}{x}}{\ln(1+2x)} &= \lim_{x\to 0^+} \sin\left( \frac{1}{x} \right) \frac{x^2 \cdot 2x}{\ln(1+2x) \cdot 2x}\\
&= \lim_{x\to 0^+} \sin\left( \frac{1}{x} \right) \frac{x}{2}\\
&= 0\\
\end{align}
