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Tìm các giới hạn sau
a) \(\mathop {\lim }\limits_{x \to {2^ – }} {{{x^2} – 4} \over {\sqrt {\left( {{x^2} + 1} \right)\left( {2 – x} \right)} }}\) b) \(\mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ – }} {{{x^2} + 3x + 2} \over {\left| {x + 1} \right|}}\)
c) \(\mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ + }} {{{x^2} + 3x + 2} \over {\left| {x + 1} \right|}}\) d) \(\mathop {\lim }\limits_{x \to {1^ + }} {{{x^3} – 1} \over {\sqrt {{x^2} – 1} }}.\)
a) \(\mathop {\lim }\limits_{x \to {2^ – }} {{ – \left( {x + 2} \right)\sqrt {2 – x} } \over {\sqrt {{x^2} + 1} }} = 0\);
b) Với \(x < – 1,\) ta có \(x + 1 < 0.\) Do đó \(\left| {x + 1} \right| = – x – 1\) và
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\(\mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ – }} {{{x^2} + 3x + 2} \over {\left| {x + 1} \right|}} = \mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ – }} \left( { – x – 2} \right) = – 1.\)
c) \(\mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ + }} {{\left( {x + 1} \right)\left( {x + 2} \right)} \over {x + 1}} = \mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ + }} \left( {x + 2} \right) = – 1 + 2 = 1\);
d) \(\mathop {\lim }\limits_{x \to {1^ + }} {{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)} \over {\sqrt {{x^2} – 1} }} = \mathop {\lim }\limits_{x \to {1^ + }} {{\sqrt {x – 1} \left( {{x^2} + x + 1} \right)} \over {\sqrt {x + 1} }} = 0\).