Let f(x)=(cos(2x)-cos(x))/x2.
f(0.1)=f(-0.1)=-1.493759,
f(0.01)=f(-0.01)=-1.499938,
f(0.001)=f(-0.001)=-1.499999,
f(0.0001)=f(-0.0001)=-1.500000.
f(x)=(2cos2(x)-cos(x)-1)/x2=(2cos(x)+1)(cos(x)-1)/x2.
When x is very small, cos(x)≃1-x2/2, so cos(x)-1=-x2/2; (cos(x)-1)/x2≃-½. This is the limit as x→0.
Limit as x→0 of 2cos(x)+1=2cos(0)+1=3, therefore limit f(x) as x→0 is (3)(-½)=-3/2=-1.5 as confirmed above by plugging in increasingly smaller values for x.