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 : \(\mathop {\lim }\limits_{x \to {0^ + }} {{x + 2\sqrt x } \over {x - \sqrt x }} = \mathop {\lim }\limits_{x \to {0^ + }} {{\sqrt x + 2} \over {\sqrt x - 1}} = {2 \over { - 1}} = - 2\)
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b. Với \(x < 2\), ta có : \({{4 - {x^2}} \over {\sqrt {2 - x} }} = {{\left( {2 - x} \right)\left( {2 + x} \right)} \over {\sqrt {2 - x} }} = \left( {x + 2} \right)\sqrt {2 - x} \)
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 \(\mathop {\lim }\limits_{x \to {3^ - }} {{\sqrt {{x^2} - 7x + 12} } \over {\sqrt {9 - {x^2}} }} = {1 \over {\sqrt 6 }} = {{\sqrt 6 } \over 6}\)