[tex] l =\displaystyle\lim\limits_{x\to \infty} \dfrac{1}{x}\int_0^x \dfrac{1}{2+\cos t}\, dt \\ \\ -1\leq \cos t\leq 1 \Rightarrow \dfrac{1}{2+\cos t} > 0 \\ \\ \Rightarrow\int_0^x \dfrac{1}{2+\cos t}\, dt \to \infty \\\\l = \lim\limits_{x\to \infty} \dfrac{\int_0^x \dfrac{1}{2+\cos t}\, dt}{x} = \dfrac{\infty}{\infty} \\ \\ \text{Aplicam L'Hopital.}\\\\ \dfrac{d}{dx}\int_0^x f(t)\, dt = f(x) \\ \\ l = \lim\limits_{x\to \infty} \dfrac{\dfrac{1}{2+\cos x}}{1} \\ \\ \text{Nu exista deoarece } \cos x\\\text{oscilează între -1 si 1.} [/tex]
