[tex]-1\leq cost\leq1 \Big|+2\Rightarrow 1 \leq cost+2\leq 3\Big|^{-1} \Rightarrow \dfrac{1}{1} \geq \dfrac{1}{cost+2} \geq \dfrac{1}{3} \Rightarrow \\ \\ \Rightarrow \dfrac{1}{3} \leq \dfrac{1}{cost+2} \leq 1 \Big|($integram$) \Rightarrow \int\limits^x_0 {\dfrac{1}{3}} \, dt \leq \int\limits^x_0 {\dfrac{1}{cost+2}} \, dt \leq \int\limits^x_0 {1} \, dt \Rightarrow[/tex]
[tex] \Rightarrow \dfrac{t}{3}\Big|_0^x \leq \int\limits^x_0 {\dfrac{1}{cost+2}} \, dt \leq t\big|_0^x \Rightarrow \dfrac{x}{3} -0 \leq \int\limits^x_0 {\dfrac{1}{cost+2}} \, dt \leq x-0 \Rightarrow \\ \\ \Rightarrow \dfrac{x}{3} \leq \int\limits^x_0 {\dfrac{dt}{cost+2}}\leq x [/tex]
[tex]\Rightarrow \underset{x\rightarrow \infty}{lim} $ $ \dfrac{1}{x}\cdot \int\limits^x_0 {\dfrac{dt}{cost+2}} \, = \underset{x\rightarrow \infty}{lim} $ $ \dfrac{1}{x}\cdot x = \underset{x\rightarrow \infty}{lim} $ $ 1 = 1[/tex]
