I assume you're familiar with the limit
[tex]\displaystyle\lim_{x\to0}\frac{\sin ax}{ax}=1[/tex]
for [tex]a\neq0[/tex]. We write the given limit in this form:
[tex]\displaystyle\lim_{x\to0}\frac{\sin(-2x)}x=-2\cdot\lim_{x\to0}\frac{\sin(-2x)}{-2x}[/tex]
The limit on the RHS is 1, so we're left with -2.
