Salut, am si eu nevoie de ajutor la ex 73 din poza( daca se poate explicatie pas cu pas, va rog ) .

Răspuns:

lim2nπ(3n+1)=2π/3≠kπ   unde  k∈Z

sin2π/3=√3/2<1=>

lim sinⁿ2π/3=0

Explicație pas cu pas:

[tex]\lim\limits_{n\to \infty} \Big(\sin^n\dfrac{2n\pi}{3n+1}\Big) = \lim\limits_{n\to \infty} \Big(\sin^n\dfrac{2\pi\cdot n}{3\cdot n+1}\Big)  = \lim\limits_{n\to \infty} \Big(\sin^n\dfrac{2\pi}{3}\Big)  = \\ \\ =\lim\limits_{n\to \infty} \Big(\sin^n\Big(\pi -\dfrac{\pi}{3}\Big) \Big) = \lim\limits_{n\to \infty} \Big(\sin^n\dfrac{\pi}{3}\Big)   =\lim\limits_{n\to \infty} \Big(\sin\dfrac{\pi}{3}\Big) ^n = \\ \\ = \lim\limits_{n\to \infty} \Big(\dfrac{\sqrt 3}{2}\Big)^n = 0 \\ \\ \dfrac{\sqrt 3}{2} < 1[/tex]