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Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to {0^ + }} {{\sqrt {{x^2} + x} – \sqrt x } \over {{x^2}}}\)
b. \(\mathop {\lim }\limits_{x \to {1^ – }} x{{\sqrt {1 – x} } \over {2\sqrt {1 – x} + 1 – x}}\)
c. \(\mathop {\lim }\limits_{x \to {3^ – }} {{3 – x} \over {\sqrt {27 – {x^3}} }}\)
d. \(\mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {{x^3} – 8} } \over {{x^2} – 2x}}\)
a.
\(\eqalign{
& \mathop {\lim }\limits_{x \to {0^ + }} {{\sqrt {{x^2} + x} – \sqrt x } \over {{x^2}}} = \mathop {\lim }\limits_{x \to {0^ + }} {x^2 \over {{x^2}\left( {\sqrt {{x^2} + x} + \sqrt x } \right)}} \cr
& = \mathop {\lim }\limits_{x \to {0^ + }} {1 \over {\left( {\sqrt {{x^2} + x} + \sqrt x } \right)}} = + \infty \cr} \)
b. \(\mathop {\lim }\limits_{x \to {1^ – }} x{{\sqrt {1 – x} } \over {2\sqrt {1 – x} + 1 – x}} = \mathop {\lim }\limits_{x \to {1^ – }} {x \over {2 + \sqrt {1 – x} }} = {1 \over 2}\)
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c.
\(\eqalign{
& \mathop {\lim }\limits_{x \to {3^ – }} {{3 – x} \over {\sqrt {27 – {x^3}} }} = \mathop {\lim }\limits_{x \to {3^ – }} {{{{\left( {\sqrt {3 – x} } \right)}^2}} \over {\sqrt {\left( {3 – x} \right)\left( {{x^2} + 3x + 9} \right)} }} \cr
& = \mathop {\lim }\limits_{x \to {3^ – }} {{\sqrt {3 – x} } \over {\sqrt {{x^2} + 3x + 9} }} = 0 \cr} \)
d.
\(\eqalign{
& \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {{x^3} – 8} } \over {{x^2} – 2x}} = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {\left( {x – 2} \right)\left( {{x^2} + 2x + 4} \right)} } \over {x\left( {x – 2} \right)}} \cr
& = \mathop {\lim }\limits_{x \to {2^ + }} {1 \over x}\sqrt {{{{x^2} + 2x + 4} \over {x – 2}}} = + \infty \cr} \)
Vì
\(\eqalign{
& \mathop {\lim }\limits_{x \to {2^ + }} \sqrt {{x^2} + 2x + 4} = 2\sqrt 3 \cr
& \mathop {\lim }\limits_{x \to {2^ + }} x\sqrt {x – 2} = 0;\,x\sqrt {x – 2} > 0\,\forall x > 2 \cr} \)