Bài tập 30 trang 159 SGK Toán 11 NC
Bài tập 30 trang 159 SGK Toán 11 NC
Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to \sqrt 3 } \left| {{x^2} - 8} \right|\)
b) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + x + 1}}{{{x^2} + 2x}}\)
c) \(\mathop {\lim }\limits_{x \to - 1} \sqrt {\frac{{{x^3}}}{{{x^2} - 3}}} \)
d) \(\mathop {\lim }\limits_{x \to 3} \sqrt[3]{{\frac{{2x\left( {x + 1} \right)}}{{{x^2} - 6}}}}\)
e) \(\mathop {\lim }\limits_{x \to - 2} \frac{{\sqrt {1 - {x^3}} - 3x}}{{2{x^2} + x - 3}}\)
f) \(\mathop {\lim }\limits_{x \to - 2} \frac{{2\left| {x + 1} \right| - 5\sqrt {{x^2} - 3} }}{{2x + 3}}\)
a) \(\mathop {\lim }\limits_{x \to \sqrt 3 } \left| {{x^2} - 8} \right| = \left| {{{\left( {\sqrt 3 } \right)}^2} - 8} \right| = 5\)
b) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + x + 1}}{{{x^2} + 2x}} = \frac{{{2^2} + 2 + 1}}{{{2^2} + 2.2}} = \frac{7}{8}\)
c) \(\mathop {\lim }\limits_{x \to - 1} \sqrt {\frac{{{x^3}}}{{{x^2} - 3}}} = \sqrt {\frac{1}{2}} = \frac{{\sqrt 2 }}{2}\)
d) \(\mathop {\lim }\limits_{x \to 3} \sqrt[3]{{\frac{{2x\left( {x + 1} \right)}}{{{x^2} - 6}}}} = \sqrt[3]{{\frac{{24}}{3}}} = 2\)
e) \(\mathop {\lim }\limits_{x \to - 2} \frac{{\sqrt {1 - {x^3}} - 3x}}{{2{x^2} + x - 3}} = \frac{{3 + 6}}{{8 - 5}} = 3\)
f) \(\mathop {\lim }\limits_{x \to - 2} \frac{{2\left| {x + 1} \right| - 5\sqrt {{x^2} - 3} }}{{2x + 3}} = \frac{{2 - 5}}{{ - 4 + 3}} = 3\)
-- Mod Toán 11
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