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Tìm các giới hạn sau
a) \(\mathop {\lim }\limits_{x \to – 1} {{2x + 1} \over {{x^2} – 3x + 4}}\) b) \(\mathop {\lim }\limits_{x \to 2} \sqrt {{{{x^3} + 2x + 3} \over {{x^2} + 5}}} \)
c) \(\mathop {\lim }\limits_{x \to – 2} {{{x^3} – {x^2} – x + 10} \over {{x^2} + 3x + 2}}\) d) \(\mathop {\lim }\limits_{x \to – 3} \left| {{{9 – {x^2}} \over {2{x^2} + 7x + 3}}} \right|.\)
a) \( – {1 \over 8};\) b) \({{\sqrt {15} } \over 3};\)
c) \({{{x^3} – {x^2} – x + 10} \over {{x^2} + 3x + 2}} = {{\left( {x + 2} \right)\left( {{x^2} – 3x + 5} \right)} \over {\left( {x + 1} \right)\left( {x + 2} \right)}} = {{{x^2} – 3x + 5} \over {x + 1}}\) với mọi \(x \ne -2.\) Do đó
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\(\mathop {\lim }\limits_{x \to – 2} {{{x^3} – {x^2} – x + 10} \over {{x^2} + 3x + 2}} = \mathop {\lim }\limits_{x \to – 2} {{{x^2} – 3x + 5} \over {x + 1}} = – 15;\)
d)
\(\eqalign{
& {{9 – {x^2}} \over {2{x^2} + 7x + 3}} = {{\left( {3 – x} \right)\left( {3 + x} \right)} \over {\left( {2x + 1} \right)\left( {x + 3} \right)}} = {{3 – x} \over {2x + 1}} \cr
& \mathop {\lim }\limits_{x \to – 3} \left| {{{9 – {x^2}} \over {2{x^2} + 7x + 3}}} \right| = \mathop {\lim }\limits_{x \to – 3} \left| {{{3 – x} \over {2x + 1}}} \right| \cr&= \left| {{{3 – \left( { – 3} \right)} \over {2.\left( { – 3} \right) + 1}}} \right| = {6 \over 5} \cr} \)