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.