Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to - 1} \sqrt {{{\left( {{x^2} + 1} \right)\left( {1 - 2x} \right)} \over {{x^2} + x + 1}}} \)
b) \(\mathop {\lim }\limits_{x \to 11} \root 3 \of {{{{x^2} - 9x - 22} \over {\left( {x - 11} \right)\left( {{x^2} - 3x + 16} \right)}}} \)
c) \(\mathop {\lim }\limits_{x \to + \infty } \sqrt {2{x^3} - {x^2} + 10} \)
d) \(\mathop {\lim }\limits_{x \to {{\left( { - 4} \right)}^ - }} \left( {{2 \over {{x^2} + 3x - 4}} - {3 \over {x + 4}}} \right).\)
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a) \(\sqrt 6 ;\) b) \({1 \over 2};\) c) \( + \infty ;\)
d) \({2 \over {{x^2} + 3x - 4}} - {3 \over {x + 4}} = {2 \over {\left( {x - 1} \right)\left( {x + 4} \right)}} - {3 \over {x + 4}}\)
\( = {{5 - 3x} \over {\left( {x - 1} \right)\left( {x + 4} \right)}} = {1 \over {x + 4}}.{{5 - 3x} \over {x - 1}}.\)
Vì \(\mathop {\lim }\limits_{x \to {{\left( { - 4} \right)}^ - }} {1 \over {x + 4}} = - \infty \) và \(\mathop {\lim }\limits_{x \to {{\left( { - 4} \right)}^ - }} {{5 - 3x} \over {x - 1}} = - {{17} \over 5} < 0\) nên
\(\mathop {\lim }\limits_{x \to {{\left( { - 4} \right)}^ - }} \left( {{2 \over {{x^2} + 3x - 4}} - {3 \over {x + 4}}} \right) = + \infty .\)