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Tìm các giới hạn sau
a) \(\mathop {\lim }\limits_{x \to 3} {\left( {3 – 4x} \right)^2}\) b) \(\mathop {\lim }\limits_{x \to – 1} {{{x^2} + x + 1} \over {2{x^5} + 3}}\)
c) \(\mathop {\lim }\limits_{x \to 1} {{{x^2}\left( {2x – 1} \right)} \over {{x^4} + x + 1}}\) d) \(\mathop {\lim }\limits_{x \to 2} \root 3 \of {{{{x^2} – x + 1} \over {{x^2} + 2x}}} \)
e) \(\mathop {\lim }\limits_{x \to 3} \sqrt {{{9{x^2} – x} \over {\left( {2x – 1} \right)\left( {{x^4} – 3} \right)}}} \) f) \(\mathop {\lim }\limits_{x \to 0} {{1 – {1 \over x}} \over {1 + {1 \over x}}}\)
g) \(\mathop {\lim }\limits_{x \to + \infty } \left| {{{ – {x^2} – x + 6} \over {{x^2} + 3x}}} \right|\) h) \(\mathop {\lim }\limits_{x \to – 2} {{{{\left( {{x^2} – x + 6} \right)}^2}} \over {{x^3} + 2{x^2}}}\)
a) 81; b) 1; c) \({1 \over 3};\)
d) \({{\root 3 \of 3 } \over 2};\) e) \({{\sqrt 5 } \over 5};\)
f) Với mọi \(x \ne 0,\) ta có
\({{1 – {1 \over x}} \over {1 + {1 \over x}}} = {{x – 1} \over {x + 1}}\)
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Do đó
\(\mathop {\lim }\limits_{x \to 0} {{1 – {1 \over x}} \over {1 + {1 \over x}}} = \mathop {\lim }\limits_{x \to 0} {{x – 1} \over {x + 1}} = – 1;\)
g) \({{ – {x^2} – x + 6} \over {{x^2} + 3x}} = {{2 – x} \over x}\) với mọi \(x \ne -3\)
\(\mathop {\lim }\limits_{x \to – 3} {{ – {x^2} – x + 6} \over {{x^2} + 3x}} = \mathop {\lim }\limits_{x \to – 3} {{2 – x} \over x} = -{5 \over 3}.\) Do đó
\(\mathop {\lim }\limits_{x \to – 3} \left| {{{ – {x^2} – x + 6} \over {{x^2} + 3x}}} \right| = \left| { – {5 \over 3}} \right| = {5 \over 3}.\)
h) \({{{{\left( {{x^2} – x – 6} \right)}^2}} \over {{x^3} + 2{x^2}}} = {{{{\left( {x – 3} \right)}^2}\left( {x + 2} \right)} \over {{x^2}}}\) với mọi \(x \ne 2\)
\(\mathop {\lim }\limits_{x \to – 2} {{{{\left( {{x^2} – x – 6} \right)}^2}} \over {{x^3} + 2{x^2}}} = 0\)