Step 1: Understand the pieces.
We study $f(x)=\dfrac{([x]+|x|-x)x}{\sin|x|}$ near $0$. Recall $[x]$ is the greatest integer function: for $x\to0^+$, $[x]=0$; for $x\to0^-$, $[x]=-1$.
Step 2: Right-hand limit.
For $x>0$, $|x|=x$, so the bracket is $[x]+|x|-x=0+x-x=0$. Hence the numerator is $0$ and \[ \beta=\lim_{x\to0^+}f(x)=0. \]
Step 3: Set up the left side.
For $x<0$, $[x]=-1$ and $|x|=-x$, so the bracket is $-1+(-x)-x=-1-2x$, and $\sin|x|=\sin(-x)=-\sin x$.
Step 4: Write the left expression.
\[ f(x)=\frac{(-1-2x)\,x}{-\sin x}=\frac{(1+2x)\,x}{\sin x}. \]
Step 5: Take the limit.
As $x\to0^-$, $\dfrac{x}{\sin x}\to1$ and $(1+2x)\to1$, so $\alpha=\lim_{x\to0^-}f(x)=1$.
Step 6: Compare the options.
With $\alpha=1$ and $\beta=0$, we get $\alpha-\beta=1$, which is option (2).
\[ \boxed{\alpha-\beta=1\ \text{(option 2)}} \]