How do I solve this using the limit chain rule?

Question

The question is to find the limit: $$\lim\limits_{t \rightarrow 0}(\cos 2t)^{\frac{1}{t^2}}$$

The answer is $\frac{1}{e^2}$, however I do not know how or why. Could someone please explain and maybe do the solution for this?

I tried some workings and got to e^((1/t^2)*ln(cos2t))

Answer 1

Hint

$$y=\sqrt[t^2]{\cos (2 t)}\implies \log(y)=\frac{\log(\cos(2t))}{t^2}$$ Now, use l'Hospital rule or Taylor series. When done, remember that $y=e^{\log(y)}$.

Answer 2

Hint:

$$\lim_{t\to 0}\left(\cos 2t\right)^{1/t^2}=\lim_{t\to 0}\left(1+\cos 2t-1\right)^{1/t^2}=\exp\left(\lim_{t\to 0}\frac{\cos 2t-1}{t^2}\right)$$

Answer 3

A possible approach uses

• $\lim_{y\to 0}(1-y)^{\frac 1y} = \frac 1e$

$$(\cos 2t)^{\frac 1{t^2}}= \left(\underbrace{\left(1-2\sin^2t\right)^{\frac 1{2\sin^2t}}}_{\stackrel{t\to 0}{\longrightarrow}\frac 1e}\right)^{\underbrace{\frac{2\sin^2 t}{t^2}}_{\stackrel{t\to 0}{\longrightarrow}2}}\stackrel{t\to 0}{\longrightarrow} \frac 1{e^2}$$