Вычислить предел

(1-cos4x) / 2xtg2x=2sin²2x:2xsin2x/cos2x = (2sin²2xcos2x) / 2xsin2x = (2sin2xcos2x) / 2x=

= (sin4x) / 2x

lim (1-cos4x) / 2xtg2x=lim (sin4x) / 2x=lim (sin4x) ' / (2x) '=lim (4cos4x) / 2=lim2cos4x=

=2cos0=2*1=2

Lim (1-cos (4*x)) / (2*x*tg (2*x)) = Lim ((1-cos (4*x)) ') / ((2*x*tg (2*x)) ') =

=Lim (4*sin (4*x)) / (2*tg (2*x) + 4*x / ((cos (2*x)) ^2) =

=Lim ((4*sin (4*x)) ') / ((2*tg (2*x) + 4*x / ((cos (2*x)) ^2) ') =

=Lim (16*cos (4*x)) / (4 / ((cos (2*x)) ^2) + (4 * (cos (2*x)) ^2) - 4*x*2*cos (2*x) * 2) / ((cos (2*x)) ^4)) =

=16*cos (0) / (4 / (cos (0) ^2) + (4 * (cos (0)) ^2-0) / (cos (0)) ^4) = 16 / (4+4) = 2.