Tìm các giới hạn sau :
LG a
\(\mathop {\lim }\limits_{x \to {2^ + }} {{2x + 1} \over {x - 2}}\)
Lời giải chi tiết:
\(\eqalign{
& \mathop {\lim }\limits_{x \to {2^ + }} {{2x + 1} \over {x - 2}} = + \infty \cr
& \text{vì }\,\mathop {\lim }\limits_{x \to {2^ + }} \left( {2x + 1} \right) = 5,\cr &\mathop {\lim }\limits_{x \to {2^ + }} \left( {x - 2} \right) = 0\,\text{ và }\,x - 2 > 0,\forall x > 2 \cr} \)
LG b
\(\mathop {\lim }\limits_{x \to {2^ - }} {{2x + 1} \over {x - 2}}\)
Lời giải chi tiết:
\(\eqalign{
& \mathop {\lim }\limits_{x \to {2^ - }} {{2x + 1} \over {x - 2}} = - \infty \cr
& \text{vì }\,\mathop {\lim }\limits_{x \to {2^ - }} \left( {2x + 1} \right) = 5,\cr &\mathop {\lim }\limits_{x \to {2^ - }} \left( {x - 2} \right) = 0\,\text{ và }\,x - 2 < 0,\forall x < 2 \cr} \)
LG c
\(\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} - {1 \over {{x^2}}}} \right)\)
Lời giải chi tiết:
\(\eqalign{
& \mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} - {1 \over {{x^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} {{x - 1} \over {{x^2}}} = - \infty \cr
& \text{vì }\,\mathop {\lim }\limits_{x \to 0} \left( {x - 1} \right) = - 1 < 0\cr &\text{ và }\,\mathop {\lim }\limits_{x \to 0} {x^2} = 0,{x^2} > 0\;\forall x \ne 0. \cr} \)
LG d
\(\mathop {\lim }\limits_{x \to {2^ - }} \left( {{1 \over {x - 2}} - {1 \over {{x^2} - 4}}} \right)\)
Lời giải chi tiết:
\(\eqalign{
& \mathop {\lim }\limits_{x \to {2^ - }} \left( {{1 \over {x - 2}} - {1 \over {{x^2} - 4}}} \right) \cr &= \mathop {\lim }\limits_{x \to {2^ - }} {{x + 2 - 1} \over {{x^2} - 4}} = \mathop {\lim }\limits_{x \to {2^ - }} {{x + 1} \over {{x^2} - 4}} \cr &= - \infty \cr
& \text{vì }\,\mathop {\lim }\limits_{x \to {2^ - }} \left( {x + 1} \right) = 3,\cr &\mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} - 4} \right) = 0\,\text{ và }\,{x^2} - 4 < 0\cr &\text{ với }\, - 2 < x < 2 \cr} \)
