(cos 2x − cos x)/x2 when x = 0.1,0.01,0.001,0.0001,−0.1,−0.01,−0.001,−0.0001

Question

round answers to 6 decimal places

Answer 1

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.