lim x→0 (cos(x)/cos(2*x))^(1/(x^2))

Question

If i'm not wrong, I could do (cosx/cos2x)^(1/x^2) = (cos2x/cosx)^(2x), which gives me 1 for the limit. Wolfram alpha thinks however, that it is supposed to be e^(3/2). I don't get it. Thanks for your help.

Answer 1

Sorry, but I'm afraid your assumption is wrong, and Wolfram is right.

 

All the limits are taken as x -> 0.

Lim ((cos(x)/cos(2x))^(1/x^2)

= Lim exp((1/x^2)*ln(cos(x)/cos(2x)))

= exp(Lim (1/x^2)*ln(cos(x)/cos(2x)))

= exp(Lim (1/x^2)*(ln(cos(x)) – ln(cos(2x)))

= exp(Lim (1/x^2)*ln(cos(x)) – (1/x^2)*ln(cos(2x)))

Using l’Hôpital’s rule on Lim (1/x^2)*ln(cos(x)) and Lim (1/x^2)*ln(cos(2x)), we get

Lim (1/x^2)*ln(cos(x)) = -sin(x)/(2x.cos(x))

Lim (1/x^2)*ln(cos(2x)) = -2sin(2x)/(2x.cos(2x))

From which we continue with

= exp(Lim (1/2)(-sin(x).cos(2x) + 2sin(2x).cos(x)) / (x.cos(2x).cos(x)))

= exp((1/2)*Lim (-sin(x).cos(2x) / (x.cos(2x).cos(x)) + 2*Lim(sin(2x).cos(x) / (x.cos(2x).cos(x)))

= exp((1/2)*Lim (-sin(x) / (x.cos(x)) + 2*Lim(sin(2x) / (x.cos(2x)))

= exp((1/2)*[Lim (-sin(x) / (x)*Lim(1/cos(x)) + 2*Lim(2sin(x).cos(x) / (x.cos(2x))])

= exp((1/2)*[(-1)*(1) + 2*Lim(2sin(x)/x)*Lim(cos(x) / (2cos^2(x) – 1)])

= exp((1/2)*[-1 + 2*2*(1) / (2 – 1)])

= exp((1/2)*[-1 + 4])

= exp(3/2)