Cho hàm số f(x) thoả mãn \(\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = 2\,022\). Tính \(\mathop {\lim }\limits_{x \to + \infty } \frac{{xf\left( x \right)}}{{x + 1}}\).  

Ta có \(\mathop {\lim }\limits_{x \to + \infty } \frac{{xf\left( x \right)}}{{x + 1}}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{f\left( x \right)}}{{1 + \frac{1}{x}}} = \frac{{\mathop {\lim }\limits_{x \to + \infty } f\left( x \right)}}{{\mathop {\lim }\limits_{x \to + \infty } \left( {1 + \frac{1}{x}} \right)}}\)

\( = \frac{{\mathop {\lim }\limits_{x \to + \infty } f\left( x \right)}}{{\mathop {\lim }\limits_{x \to + \infty } 1 + \mathop {\lim }\limits_{x \to + \infty } \frac{1}{x}}} = \frac{{2022}}{{1 + 0}} = 2022\).

Vậy \(\mathop {\lim }\limits_{x \to + \infty } \frac{{xf\left( x \right)}}{{x + 1}} = 2022\).