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Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} + {1 \over {{x^2}}}} \right)\)
b. \(\mathop {\lim }\limits_{x \to – 2} {{{x^3} + 8} \over {x + 2}}\)
c. \(\mathop {\lim }\limits_{x \to 9} {{3 – \sqrt x } \over {9 – x}}\)
d. \(\mathop {\lim }\limits_{x \to 0} {{2 – \sqrt {4 – x} } \over x}\)
e. \(\mathop {\lim }\limits_{x \to + \infty } {{{x^4} – {x^3} + 11} \over {2x – 7}}\)
f. \(\mathop {\lim }\limits_{x \to – \infty } {{\sqrt {{x^4} + 4} } \over {x + 4}}\)
a. \(\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} + {1 \over {{x^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} {{x + 1} \over {{x^2}}} = + \infty \)
vì \(\mathop {\lim }\limits_{x \to 0} \left( {x + 1} \right) = 1 > 0,\mathop {\lim }\limits_{x \to 0} {x^2} = 0\,\text{ và }\,{x^2} > 0,\forall x \ne 0\)
b.
\(\eqalign{
& \mathop {\lim }\limits_{x \to – 2} {{{x^3} + 8} \over {x + 2}} = \mathop {\lim }\limits_{x \to – 2} {{\left( {x + 2} \right)\left( {{x^2} – 2x + 4} \right)} \over {x + 2}} \cr
& = \mathop {\lim }\limits_{x \to – 2} \left( {{x^2} – 2x + 4} \right) = 12 \cr} \)
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c. \(\mathop {\lim }\limits_{x \to 9} {{3 – \sqrt x } \over {9 – x}} = \mathop {\lim }\limits_{x \to 9} {1 \over {3 + \sqrt x }} = {1 \over 6}\)
d.
\(\eqalign{
& \mathop {\lim }\limits_{x \to 0} {{2 – \sqrt {4 – x} } \over x} = \mathop {\lim }\limits_{x \to 0} {{4 – \left( {4 – x} \right)} \over {x\left( {2 + \sqrt {4 – x} } \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} {1 \over {2 + \sqrt {4 – x} }} = {1 \over 4} \cr} \)
e.
\(\mathop {\lim }\limits_{x \to + \infty } {{{x^4} – {x^3} + 11} \over {2x – 7}}\)
\(=\mathop {\lim }\limits_{x \to + \infty } {{{x^3} – {x^2} + {{11} \over x}} \over {2 – {7 \over x}}} = + \infty \)
f. Với \(x < 0\), ta có : \({{\sqrt {{x^4} + 4} } \over {x + 4}} = {{{x^2}\sqrt {1 + {4 \over {{x^4}}}} } \over {x + 4}} = {{x\sqrt {1 + {4 \over {{x^2}}}} } \over {1 + {4 \over x}}}\)
vì \(\mathop {\lim }\limits_{x \to – \infty } x\sqrt {1 + {4 \over {{x^4}}}} = – \infty \,\text{ và }\,\mathop {\lim }\limits_{x \to – \infty } \left( {1 + {4 \over x}} \right) = 1\)
nên \(\mathop {\lim }\limits_{x \to – \infty } {{\sqrt {{x^4} + 4} } \over {x + 4}} = – \infty \)