[tex] L = \lim\limits_{x\to +\infty} \dfrac{3x-\sin x}{x + \sin x}[/tex]
[tex]-1 \leq \sin x \leq 1 \\ \\ -\dfrac{1}{x} \leq \dfrac{\sin x}{x} \leq \dfrac{1}{x} \\ \\ \lim\limits_{x \to +\infty } \left(-\dfrac{1}{x}\right) \leq \lim\limits_{x \to +\infty} \dfrac{\sin x}{x} \leq \lim\limits_{x \to +\infty } \dfrac{1}{x} \\ \\ 0 \leq \lim\limits_{x \to +\infty }\dfrac{\sin x}{x} \leq 0 \\ \\ \Rightarrow \boxed{\lim\limits_{x \to +\infty } \dfrac{\sin x}{x} = 0} \Rightarrow \boxed{\lim\limits_{u(x) \to + \infty } \dfrac{\sin \Big(u(x)\Big)}{u(x)} = 0} [/tex]
[tex]L = \lim\limits_{x \to +\infty }\dfrac{x\cdot \Big(3-\dfrac{\sin x}{x}\Big)}{x\cdot\Big (1+\dfrac{\sin x}{x}\Big)} = \lim\limits_{x \to +\infty } \dfrac{3-\dfrac{\sin x}{x}}{1+\dfrac{\sin x}{x}} = \dfrac{3- \lim\limits_{x\to +\infty}\dfrac{\sin x}{x}}{1+\lim\limits_{x\to +\infty } \dfrac{\sin x}{x}} = \\ \\ = \dfrac{3-0}{1+0} = 3[/tex]
