Tìm các giới hạn sau
a) lim 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
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{{1 - {1 \over x}} \over {1 + {1 \over x}}} = {{x - 1} \over {x + 1}}
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