You chose the poorer way to turn the $0\cdot\infty$ form into one to which l’Hospital’s rule applies.
$$\begin{align*}
\lim_{x\to 0^+}\ln x\sin x&=\lim_{x\to 0^+}\frac{\sin x}{1/\ln x}\\\\
&=\lim_{x\to 0^+}\frac{\cos x}{-(\ln x)^{-2}\cdot\frac1x}\\\\
&=-\lim_{x\to 0^+}\frac{(\ln x)^2\cos x}{1/x}\;.
\end{align*}$$
At this point you can try try applying l’Hospital’s rule again, but it’s clear that the numerator is going to be fairly messy. A better idea is to notice that $\lim\limits_{x\to 0^+}\cos x= 1$, so that
$$-\lim_{x\to 0^+}\frac{(\ln x)^2\cos x}{1/x}=-\left(\lim_{x\to 0^+}\frac{(\ln x)^2}{1/x}\right)\lim_{x\to 0^+}\cos x=-\lim_{x\to 0^+}\frac{(\ln x)^2}{1/x}\;;$$ this gets rid of the trig function. Now
$$\begin{align*}
-\lim_{x\to 0^+}\frac{(\ln x)^2}{1/x}&=-\lim_{x\to 0^+}\frac{\frac2x\ln x}{-1/x^2}\\\\
&=2\lim_{x\to 0^+}\frac{\ln x}{1/x}\\\\
&=2\lim_{x\to 0^+}\frac{1/x}{-1/x^2}\\\\
&=-2\lim_{x\to 0^+}x\\\\
&=0\;.
\end{align*}$$
It’s always a good idea to keep your eyes open for factors with known finite, non-zero limits, like the $\cos x$ above: generally speaking, it’s a good idea to simplify as much as possible the expression whose limit you’re taking.