Bài tập 4.54 trang 173 SBT Toán 11

a) \(\mathop {\lim }\limits_{x \to  - 2} \frac{{x + 5}}{{{x^2} + x - 3}} = \frac{{\left( { - 2} \right) + 5}}{{{{\left( { - 2} \right)}^2} + \left( { - 2} \right) - 3}} =  - 3\)

b) \(\mathop {\lim }\limits_{x \to {3^ - }} \sqrt {{x^2} + 8x + 3}  = \sqrt {{3^2} + 8.3 + 3}  = 6\)

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

d) \(\mathop {\lim }\limits_{x \to  - 1} \frac{{2{x^3} - 5x - 4}}{{{{\left( {x + 1} \right)}^2}}} =  - \infty \)

Vì \(\mathop {\lim }\limits_{x \to  - 1} \left( {2{x^3} - 5x - 4} \right) =  - 1 < 0\) và \(\mathop {\lim }\limits_{x \to  - 1} (x + 1) = 0\) với x = −1.

-- Mod Toán 11 HỌC247