LG a
\(\mathop {\lim }\limits_{x \to 1} \left[ {{2 \over {{{\left( {x - 1} \right)}^2}}}.{{2x + 1} \over {2x - 3}}} \right]\)
Lời giải chi tiết:
Ta có: \(\mathop {\lim }\limits_{x \to 1} {2 \over {{{\left( {x - 1} \right)}^2}}} = + \infty \) vì \(\mathop {\lim }\limits_{x \to 1} {\left( {x - 1} \right)^2} = 0,{\left( {x - 1} \right)^2} > 0,\forall x \ne 1\)
\(\text{ và }\,\mathop {\lim }\limits_{x \to 1} {{2x + 1} \over {2x - 3}} = {3 \over { - 1}} = - 3 < 0\)
Do đó \(\mathop {\lim }\limits_{x \to 1} \left[ {{2 \over {{{\left( {x - 1} \right)}^2}}}.{{2x + 1} \over {2x - 3}}} \right] = - \infty \)
LG b
\(\mathop {\lim }\limits_{x \to 1} {5 \over {\left( {x - 1} \right)\left( {{x^2} - 3x + 2} \right)}}\)
Lời giải chi tiết:
\(\eqalign{
& {5 \over {\left( {x - 1} \right)\left( {{x^2} - 3x + 2} \right)}} \cr & = \frac{5}{{\left( {x - 1} \right)\left( {x - 1} \right)\left( {x - 2} \right)}}\cr &= {1 \over {{{\left( {x - 1} \right)}^2}}}.{5 \over {x - 2}} \cr
& \mathop {\lim }\limits_{x \to 1} {1 \over {{{\left( {x - 1} \right)}^2}}} = + \infty \cr &\text{ vì } \mathop {\lim }\limits_{x \to 1} {\left( {x - 1} \right)^2} = 0,{\left( {x - 1} \right)^2} > 0,\forall x \ne 1\cr &\mathop {\lim }\limits_{x \to 1} {5 \over {x - 2}} = - 5 < 0 \cr
& \text{ nên }\cr &\mathop {\lim }\limits_{x \to 1} {5 \over {\left( {x - 1} \right)\left( {{x^2} - 3x + 2} \right)}} = - \infty \cr} \)
