Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to - 2} \frac{{x + 5}}{{{x^2} + x - 3}}\)
b) \(\mathop {\lim }\limits_{x \to {3^ - }} \sqrt {{x^2} + 8x + 3} \)
c) \(\mathop {\lim }\limits_{x \to + \infty } \left( {{x^3} + 2{x^2}\sqrt x - 1} \right)\)
d)
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
-- Mod Toán 11