Câu 35 trang 163 SGK Đại số và Giải tích 11 Nâng cao

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:

\(\eqalign{
& \mathop {\lim }\limits_{x \to {2^ – }} \left( {{1 \over {x – 2}} – {1 \over {{x^2} – 4}}} \right) \cr &= \mathop {\lim }\limits_{x \to {2^ – }} {{x + 2 – 1} \over {{x^2} – 4}} = \mathop {\lim }\limits_{x \to {2^ – }} {{x + 1} \over {{x^2} – 4}} \cr &= – \infty \cr 
&  \text{vì }\,\mathop {\lim }\limits_{x \to {2^ – }} \left( {x + 1} \right) = 3,\cr &\mathop {\lim }\limits_{x \to {2^ – }} \left( {{x^2} – 4} \right) = 0\,\text{ và }\,{x^2} – 4 < 0\cr &\text{ với }\, – 2 < x < 2 \cr} \)

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