L'Hopital's is probably the easiest approach:
[tex]\displaystyle\lim_{x\to0}\frac{2x-\sin2x}{2x-\tan2x}=\lim_{x\to0}\frac{2-2\cos2x}{2-2\sec^22x}[/tex] (LHR)
[tex]\displaystyle=\lim_{x\to0}\frac{4\sin2x}{-8\sec^22x\tan2x}[/tex] (LHR)
[tex]\displaystyle=-\frac12\lim_{x\to0}\cos^32x[/tex] (simplify)
[tex]\displaystyle=-\frac12[/tex]
