Tìm giới hạn:a) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{x - 1}}{{2x + 1}}\)              &nb

* Hướng dẫn giải

a) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{x - 1}}{{2x + 1}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{1 - \frac{1}{x}}}{{2 + \frac{1}{x}}} = \frac{1}{2}\)

b) \(\mathop {\lim }\limits_{x \to 3} ({x^3} - {x^2} + 2018) = {3^3} - {3^2} + 2018 = 2036\)

c) \(\mathop {\lim }\limits_{x \to {3^ - }} \frac{{{x^2} + x + 1}}{{x - 3}} =  - \infty \)

Vì \(\mathop {\lim }\limits_{x \to {3^ - }} \left( {{x^2} + x + 1} \right) = 13 > 0,\mathop {\lim }\limits_{x \to {3^ - }} \left( {x - 3} \right) = 0\) và \(x \to {3^ - } \Rightarrow x - 3 < 0\)