Question

Find
Lim
`[cos(2x)]^(1/x^2)`
x->
`0^+`

Answer 1

`(1)/(e^2)`

Explanation:

`L=\lim_(x\rightarrow 0)(\cos 2x)^(1/x^2)\\\ln L=\ln(\lim_(x\rightarrow 0)(\cos 2x)^(1/x^2))\\\ln L=\lim_(x\rightarrow 0)\ln((\cos 2x)^(1/x^2))\\\ln L=\lim_(x\rightarrow 0)(\ln\cos 2x)/(x^2)\\\ln L=\lim_(x\rightarrow 0)(\ln\cos 2x)/(x^2)=\lim_(x\rightarrow 0)(-2\sin2x/\cos 2x)/(2x)=-2\cdot\lim_(x\rightarrow 0)(\tan 2x)/(2x)=\\=-2\cdot\underbrace{\lim_(x\rightarrow 0)(2/\cos^2 2x)/(2)}_(=1)=-2\\\\\Rightarrow\quad L=e^(-2)=(1)/(e^2)`