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