Đề bài

Cho hàm số \(f\left( x \right)\) thoả mãn \(\mathop {\lim }\limits_{x \to  + \infty } f\left( x \right) = 2022\). Tính \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{xf\left( x \right)}}{{x + 1}}\).

Lời giải chi tiết

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