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