Question

Find the limit. use l'hospital's rule if appropriate. if there is a more elementary method, consider using it. lim x→0 2x − sin(2x) 2x − tan(2x)

Answer 1

L'Hopital's is probably the easiest approach:

$\displaystyle\lim_{x\to0}\frac{2x-\sin2x}{2x-\tan2x}=\lim_{x\to0}\frac{2-2\cos2x}{2-2\sec^22x}$ (LHR)
$\displaystyle=\lim_{x\to0}\frac{4\sin2x}{-8\sec^22x\tan2x}$ (LHR)
$\displaystyle=-\frac12\lim_{x\to0}\cos^32x$ (simplify)
$\displaystyle=-\frac12$