As with any indeterminate form, you have to look at the actual behavior of the specific function, not just at the behavior of its component parts. As you say, $\ln(-x)$ blows up as $x\to-\infty$. You’re multiplying it by $\sin x+2$, which oscillates over the range $[1,3]$. If $f(x)=(\sin x+2)\ln(-x)$, at each $x<0$ you have $\ln(-x)\le f(x)\le 3\ln(-x)\;.$ As $x\to-\infty$, both $\ln(-x)$ and $3\ln(-x)$ increase without bound, and $f(x)$ is trapped between them, so it must also increase without bound: $\lim\limits_{x\to-\infty}f(x)=\infty$.
All you actually need here is the fact that $\ln(-x)$ is blowing up, since $f(x)$ is trapped above it: as $\ln(-x)$ increases without bound, $f(x)$ is forced up as well.
It would be a very different story if your function were $(x\sin x)\ln(-x)$, for instance. The oscillations in $x\sin x$ get bigger and bigger in both directions as $x\to-\infty$, so the oscillations in $(x\sin x)\ln(-x)$ do so as well: $\lim\limits_{x\to-\infty}|(x\sin x)\ln(-x)|=\infty$, but $\lim\limits_{x\to-\infty}(x\sin x)\ln(-x)$ doesn’t exist.