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Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to \sqrt 3 } \left| {{x^2} – 8} \right|\)
b. \(\mathop {\lim }\limits_{x \to 2} {{{x^2} + x + 1} \over {{x^2} + 2x}}\)
c. \(\mathop {\lim }\limits_{x \to – 1} \sqrt {{{{x^3}} \over {{x^2} – 3}}} \)
d. \(\mathop {\lim }\limits_{x \to 3} \root 3 \of {{{2x\left( {x + 1} \right)} \over {{x^2} – 6}}} \)
e. \(\mathop {\lim }\limits_{x \to – 2} {{\sqrt {1 – {x^3}} – 3x} \over {2{x^2} + x – 3}}\)
f. \(\mathop {\lim }\limits_{x \to – 2} {{2\left| {x + 1} \right| – 5\sqrt {{x^2} – 3} } \over {2x + 3}}\)
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a. \(\mathop {\lim }\limits_{x \to \sqrt 3 } \left| {{x^2} – 8} \right| = \left| {{{\left( {\sqrt 3 } \right)}^2} – 8} \right| = 5\)
b. \(\mathop {\lim }\limits_{x \to 2} {{{x^2} + x + 1} \over {{x^2} + 2x}} = {{{2^2} + 2 + 1} \over {{2^2} + 2.2}} = {7 \over 8}\)
c. \(\mathop {\lim }\limits_{x \to – 1} \sqrt {{{{x^3}} \over {{x^2} – 3}}} = \sqrt {{1 \over 2}} = {{\sqrt 2 } \over 2}\)
d. \(\mathop {\lim }\limits_{x \to 3} \root 3 \of {{{2x\left( {x + 1} \right)} \over {{x^2} – 6}}} = \root 3 \of {{{24} \over 3}} = 2\)
e. \(\mathop {\lim }\limits_{x \to – 2} {{\sqrt {1 – {x^3}} – 3x} \over {2{x^2} + x – 3}} = {{3 + 6} \over {8 – 5}} = 3\)
f. \(\mathop {\lim }\limits_{x \to – 2} {{2\left| {x + 1} \right| – 5\sqrt {{x^2} – 3} } \over {2x + 3}} = {{2 – 5} \over { – 4 + 3}} = 3\)