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Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to {0^ + }} {{x + 2\sqrt x } \over {x – \sqrt x }}\)
b. \(\mathop {\lim }\limits_{x \to {2^ – }} {{4 – {x^2}} \over {\sqrt {2 – x} }}\)
c. \(\mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ + }} {{{x^2} + 3x + 2} \over {\sqrt {{x^5} + {x^4}} }}\)
d. \(\mathop {\lim }\limits_{x \to {3^ – }} {{\sqrt {{x^2} – 7x + 12} } \over {\sqrt {9 – {x^2}} }}\)
a. Với \(x > 0\), ta có : \({{x + 2\sqrt x } \over {x – \sqrt x }} = {{\sqrt x \left( \sqrt x + 2 \right)} \over {\sqrt x \left( {\sqrt x – 1} \right)}} = {{\sqrt x + 2} \over {\sqrt x – 1}}\)
do đó : \(\mathop {\lim }\limits_{x \to {0^ + }} {{x + 2\sqrt x } \over {x – \sqrt x }} = \mathop {\lim }\limits_{x \to {0^ + }} {{\sqrt x + 2} \over {\sqrt x – 1}} = {2 \over { – 1}} = – 2\)
b. Với \(x < 2\), ta có : \({{4 – {x^2}} \over {\sqrt {2 – x} }} = {{\left( {2 – x} \right)\left( {2 + x} \right)} \over {\sqrt {2 – x} }} = \left( {x + 2} \right)\sqrt {2 – x} \)
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Do đó \(\mathop {\lim }\limits_{x \to {2^ – }} {{4 – {x^2}} \over {\sqrt {2 – x} }} = \mathop {\lim }\limits_{x \to {2^ – }} \left( {x + 2} \right)\sqrt {2 – x} = 0\)
c. Với mọi \(x > -1\)
\({{{x^2} + 3x + 2} \over {\sqrt {{x^5} + {x^4}} }} = {{\left( {x + 1} \right)\left( {x + 2} \right)} \over {{x^2}\sqrt {x + 1} }} = {{\sqrt {x + 1} \left( {x + 2} \right)} \over {{x^2}}}\)
Do đó \(\mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ + }} {{{x^2} + 3x + 2} \over {\sqrt {{x^5} + {x^4}} }} = \mathop {\lim }\limits_{x \to {{\left( { – 1} \right)}^ + }} {{\sqrt {x + 1} \left( {x + 2} \right)} \over {{x^2}}} = 0\)
d. Với \(-3 < x < 3\)
\({{\sqrt {{x^2} – 7x + 12} } \over {\sqrt {9 – {x^2}} }} = {{\sqrt {\left( {3 – x} \right)\left( {4 – x} \right)} } \over {\sqrt {\left( {3 – x} \right)\left( {3 + x} \right)} }} = {{\sqrt {4 – x} } \over {\sqrt {3 + x} }}\)
Do đó \(\mathop {\lim }\limits_{x \to {3^ – }} {{\sqrt {{x^2} – 7x + 12} } \over {\sqrt {9 – {x^2}} }} = {1 \over {\sqrt 6 }} = {{\sqrt 6 } \over 6}\)