kết quả giới hạn lim 4x^2-x 2 2x-1

Lời giải:

\[\begin{array}{l}\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {4{x^2} + x}  + 2x - 1} \right)\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{4{x^2} + x - {{(2x - 1)}^2}}}{{\sqrt {4{x^2} + x}  - 2x + 1}}\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{4{x^2} + x - 4{x^2} + 4x - 1}}{{\sqrt {4{x^2} + x}  - 2x + 1}}\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{5x}}{{ - x.\sqrt {4 + \frac{1}{x}}  - 2x + 1}}\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{5 - \frac{1}{x}}}{{ - \sqrt {4 + \frac{1}{x}}  - 2 + \frac{1}{x}}}\\ = \frac{{5 - 0}}{{ - \sqrt {4 + 0}  - 2 + 0}} = \frac{5}{{ - 4}} =  - \frac{5}{4}\end{array}\]