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a) \(\mathop {\lim }\limits_{x \to {1^ + }} {{{x^2} + 1} \over {x – 1}}\) b) \(\mathop {\lim }\limits_{x \to {1^ – }} {{{x^2} + 1} \over {x – 1}}\)
c) \(\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} {{\left| {3x + 6} \right|} \over {x + 2}}\) d) \(\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ – }} {{\left| {3x + 6} \right|} \over {x + 2}}\) .
a)
\(\eqalign{
& \mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2} + 1} \right) = 2 > 0 \cr
& \mathop {\lim }\limits_{x \to {1^ + }} \left( {x – 1} \right) > 0 \cr
& \mathop {\lim }\limits_{x \to {1^ + }} {{{x^2} + 1} \over {x – 1}} = + \infty \cr} \)
b)
\(\eqalign{
& \mathop {\lim }\limits_{x \to {1^ – }} \left( {{x^2} + 1} \right) = 2 > 0 \cr
& \mathop {\lim }\limits_{x \to {1^ – }} \left( {x – 1} \right) < 0 \cr
& \mathop {\lim }\limits_{x \to {1^ – }} {{{x^2} + 1} \over {x – 1}} = – \infty \cr} \)
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c) Với \(x > – 2,\) ta có \(3x + 6 = 3\left( {x + 2} \right) > 0.\) Do đó
\(\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} {{\left| {3x + 6} \right|} \over {x + 2}} = \mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} {{3x + 6} \over {x + 2}} = \mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} 3 = 3;\)
d) Với \(x < – 2,\) ta có \(3x + 6 = 3\left( {x + 2} \right) < 0.\) Do đó
\(\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ – }} {{\left| {3x + 6} \right|} \over {x + 2}} = \mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ – }} -{{3x + 6} \over {x + 2}} = \mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ – }}(- 3) =- 3\)