Câu 37 trang 163 SGK Đại số và Giải tích 11 Nâng cao

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:

\(\eqalign{
& {5 \over {\left( {x – 1} \right)\left( {{x^2} – 3x + 2} \right)}} \cr & = \frac{5}{{\left( {x – 1} \right)\left( {x – 1} \right)\left( {x – 2} \right)}}\cr &= {1 \over {{{\left( {x – 1} \right)}^2}}}.{5 \over {x – 2}} \cr 
& \mathop {\lim }\limits_{x \to 1} {1 \over {{{\left( {x – 1} \right)}^2}}} = + \infty \cr &\text{ vì } \mathop {\lim }\limits_{x \to 1} {\left( {x – 1} \right)^2} = 0,{\left( {x – 1} \right)^2} > 0,\forall x \ne 1\cr &\mathop {\lim }\limits_{x \to 1} {5 \over {x – 2}} = – 5 < 0 \cr 
& \text{ nên }\cr &\mathop {\lim }\limits_{x \to 1} {5 \over {\left( {x – 1} \right)\left( {{x^2} – 3x + 2} \right)}} = – \infty \cr} \)

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