Câu 38 trang 166 SGK Đại số và Giải tích 11 Nâng cao. Tìm các giới hạn sau : a. \(\mathop {\lim }\limits_{x \to 2} {{{x^3} - 8} \over {{x^2} - 4}}\) b. \(\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ + }} {{2{x^2} + 5x - 3} \over {{{\left( {x + 3} \right)}^2}}}\) c. \(\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} {{2{x^2} + 5x - 3} \over {{{\left( {x + 3} \right)}^2}}}\) d. \(\mathop {\lim }\limits_{x \to 0} {{\sqrt {{x^3} + 1} - 1} \over {{x^2} + x}}\) Giải: a. Dạng \({0 \over 0}\) ta phân tích tử và mẫu ra thừa số : \(\eqalign{ & \mathop {\lim }\limits_{x \to 2} {{{x^3} - 8} \over {{x^2} - 4}} = \mathop {\lim }\limits_{x \to 2} {{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)} \over {\left( {x - 2} \right)\left( {x + 2} \right)}} \cr & = \mathop {\lim }\limits_{x \to 2} {{{x^2} + 2x + 4} \over {x + 2}} = 3 \cr} \) b. \(\eqalign{ & \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ + }} {{2{x^2} + 5x - 3} \over {{{\left( {x + 3} \right)}^2}}} = \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ + }} {{\left( {x + 3} \right)\left( {2x - 1} \right)} \over {{{\left( {x + 3} \right)}^2}}} \cr & = \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ + }} {{2x - 1} \over {x + 3}} = - \infty \cr} \) Vì \(\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ + }} \left( {2x - 1} \right) = - 7 < 0\,\text{ và }\,\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ +}} \left( {x + 3} \right) = 0;\) \(x + 3 > 0\) c. \(\eqalign{ & \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} {{2{x^2} + 5x - 3} \over {{{\left( {x + 3} \right)}^2}}} = \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} {{\left( {x + 3} \right)\left( {2x - 1} \right)} \over {{{\left( {x + 3} \right)}^2}}} \cr & = \mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} {{2x - 1} \over {x + 3}} = + \infty \cr} \) Vì \(\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} \left( {2x - 1} \right) = - 7 < 0\,\text{ và }\,\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} \left( {x + 3} \right) = 0;\) \(x + 3 < 0\) d. \(\eqalign{ & \mathop {\lim }\limits_{x \to 0} {{\sqrt {{x^3} + 1} - 1} \over {{x^2} + x}} = \mathop {\lim }\limits_{x \to 0} {{{x^3}} \over {x\left( {x + 1} \right)\left( {\sqrt {{x^3} + 1} + 1} \right)}} \cr & = \mathop {\lim }\limits_{x \to 0} {{{x^2}} \over {\left( {x + 1} \right)\left( {\sqrt {{x^3} + 1} + 1} \right)}} = 0 \cr} \) Câu 39 trang 166 SGK Đại số và Giải tích 11 Nâng cao. Tìm các giới hạn sau : a. \(\mathop {\lim }\limits_{x \to + \infty } {{2{x^2} + x - 10} \over {9 - 3{x^3}}}\) b. \(\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {2{x^2} - 7x + 12} } \over {3\left| x \right| - 17}}\) Giải: a. \(\mathop {\lim }\limits_{x \to + \infty } {{2{x^2} + x - 10} \over {9 - 3{x^3}}} = \mathop {\lim }\limits_{x \to + \infty } {{{2 \over x} + {1 \over {{x^2}}} - {{10} \over {{x^3}}}} \over {{9 \over {{x^3}}} - 3}} = 0\) b. Với mọi \(x ≠ 0\), ta có : \({{\sqrt {2{x^2} - 7x + 12} } \over {3\left| x \right| - 17}} = {{\left| x \right|\sqrt {2 - {7 \over x} + {{12} \over {{x^2}}}} } \over {\left| x \right|\left( {3 - {{17} \over {\left| x \right|}}} \right)}} = {{\sqrt {2 - {7 \over x} + {{12} \over {{x^2}}}} } \over {3 - {{17} \over {\left| x \right|}}}}\) Do đó \(\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {2{x^2} - 7x + 12} } \over {3\left| x \right| - 17}} = {{\sqrt 2 } \over 3}\) Câu 40 trang 166 SGK Đại số và Giải tích 11 Nâng cao. Tìm các giới hạn sau : a. \(\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \left( {{x^3} + 1} \right)\sqrt {{x \over {{x^2} - 1}}} \) b. \(\mathop {\lim }\limits_{x \to + \infty } \left( {x + 2} \right)\sqrt {{{x - 1} \over {{x^3} + x}}} \) Giải: a. Dạng 0.∞ Với \(x > -1\) đủ gần -1 (\(-1 < x < 0\)) ta có : \(\eqalign{ & \left( {{x^3} + 1} \right)\sqrt {{x \over {{x^2} - 1}}} \cr &= \left( {{x^2} - x + 1} \right)\left( {x + 1} \right).\sqrt {{x \over {{x^2} - 1}}} \cr & = \left( {{x^2} - x + 1} \right)\sqrt {{{x\left( {x + 1} \right)} \over {x - 1}}} \cr & \Rightarrow \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \left( {{x^3} + 1} \right)\sqrt {{x \over {{x^2} - 1}}}\cr & \;\;= \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \left( {{x^2} - x + 1} \right)\sqrt {{{x\left( {x + 1} \right)} \over {x - 1}}} = 0 \cr} \) b. Dạng 0.∞ \(\eqalign{ & \mathop {\lim }\limits_{x \to + \infty } \left( {x + 2} \right)\sqrt {{{x - 1} \over {{x^3} + x}}} \cr &= \mathop {\lim }\limits_{x \to + \infty } \sqrt {{{{{\left( {x + 2} \right)}^2}\left( {x - 1} \right)} \over {{x^3} + x}}} \cr & = \mathop {\lim }\limits_{x \to + \infty } \sqrt {{{{{\left( {1 + {2 \over x}} \right)}^2}\left( {1 - {1 \over x}} \right)} \over {1 + {1 \over {{x^2}}}}}} = 1 \cr} \) Câu 41 trang 166 SGK Đại số và Giải tích 11 Nâng cao. Tìm các giới hạn sau : a. \(\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 1} - x} \right)\) b. \(\mathop {\lim }\limits_{x \to 1} {{\sqrt {2x - {x^2}} - 1} \over {{x^2} - x}}\) Giải: a. Dạng ∞ - ∞ \(\eqalign{ & \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 1} - x} \right) = \mathop {\lim }\limits_{x \to + \infty } {{{x^2} + 1 - {x^2}} \over {\sqrt {{x^2} + 1} + x}} \cr & = \mathop {\lim }\limits_{x \to + \infty } {1 \over {\sqrt {{x^2} + 1} + x}} = 0 \cr} \) b. Dạng \({0 \over 0}\) \(\eqalign{ & \mathop {\lim }\limits_{x \to 1} {{\sqrt {2x - {x^2}} - 1} \over {{x^2} - x}} = \mathop {\lim }\limits_{x \to 1} {{2x - {x^2} - 1} \over {x\left( {x - 1} \right)\left( {\sqrt {2x - {x^2}} + 1} \right)}} \cr & = \mathop {\lim }\limits_{x \to 1} {{ - {{\left( {x - 1} \right)}^2}} \over {x\left( {x - 1} \right)\left( {\sqrt {2x - {x^2}} + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} {{1 - x} \over {x\left( {\sqrt {2x - {x^2}} + 1} \right)}} = 0 \cr} \) Câu 42 trang 167 SGK Đại số và Giải tích 11 Nâng cao. Tìm các giới hạn sau : a. \(\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} + {1 \over {{x^2}}}} \right)\) b. \(\mathop {\lim }\limits_{x \to - 2} {{{x^3} + 8} \over {x + 2}}\) c. \(\mathop {\lim }\limits_{x \to 9} {{3 - \sqrt x } \over {9 - x}}\) d. \(\mathop {\lim }\limits_{x \to 0} {{2 - \sqrt {4 - x} } \over x}\) e. \(\mathop {\lim }\limits_{x \to + \infty } {{{x^4} - {x^3} + 11} \over {2x - 7}}\) f. \(\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^4} + 4} } \over {x + 4}}\) Giải: a. \(\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} + {1 \over {{x^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} {{x + 1} \over {{x^2}}} = + \infty \) vì \(\mathop {\lim }\limits_{x \to 0} \left( {x + 1} \right) = 1 > 0,\mathop {\lim }\limits_{x \to 0} {x^2} = 0\,\text{ và }\,{x^2} > 0,\forall x \ne 0\) b. \(\eqalign{ & \mathop {\lim }\limits_{x \to - 2} {{{x^3} + 8} \over {x + 2}} = \mathop {\lim }\limits_{x \to - 2} {{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)} \over {x + 2}} \cr & = \mathop {\lim }\limits_{x \to - 2} \left( {{x^2} - 2x + 4} \right) = 12 \cr} \) c. \(\mathop {\lim }\limits_{x \to 9} {{3 - \sqrt x } \over {9 - x}} = \mathop {\lim }\limits_{x \to 9} {1 \over {3 + \sqrt x }} = {1 \over 6}\) d. \(\eqalign{ & \mathop {\lim }\limits_{x \to 0} {{2 - \sqrt {4 - x} } \over x} = \mathop {\lim }\limits_{x \to 0} {{4 - \left( {4 - x} \right)} \over {x\left( {2 + \sqrt {4 - x} } \right)}} \cr & = \mathop {\lim }\limits_{x \to 0} {1 \over {2 + \sqrt {4 - x} }} = {1 \over 4} \cr} \) e. \(\mathop {\lim }\limits_{x \to + \infty } {{{x^4} - {x^3} + 11} \over {2x - 7}}\) \(=\mathop {\lim }\limits_{x \to + \infty } {{{x^3} - {x^2} + {{11} \over x}} \over {2 - {7 \over x}}} = + \infty \) f. Với \(x < 0\), ta có : \({{\sqrt {{x^4} + 4} } \over {x + 4}} = {{{x^2}\sqrt {1 + {4 \over {{x^4}}}} } \over {x + 4}} = {{x\sqrt {1 + {4 \over {{x^2}}}} } \over {1 + {4 \over x}}}\) vì \(\mathop {\lim }\limits_{x \to - \infty } x\sqrt {1 + {4 \over {{x^4}}}} = - \infty \,\text{ và }\,\mathop {\lim }\limits_{x \to - \infty } \left( {1 + {4 \over x}} \right) = 1\) nên \(\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^4} + 4} } \over {x + 4}} = - \infty \) Câu 43 trang 167 SGK Đại số và Giải tích 11 Nâng cao. Tìm các giới hạn sau : a. \(\mathop {\lim }\limits_{x \to - \sqrt 3 } {{{x^3} + 3\sqrt 3 } \over {3 - {x^2}}}\) b. \(\mathop {\lim }\limits_{x \to 4} {{\sqrt x - 2} \over {{x^2} - 4x}}\) c. \(\mathop {\lim }\limits_{x \to {1^ + }} {{\sqrt {x - 1} } \over {{x^2} - x}}\) d. \(\mathop {\lim }\limits_{x \to 0} {{\sqrt {{x^2} + x + 1} - 1} \over {3x}}\) Giải: a. Ta có: \({{{x^3} + 3\sqrt 3 } \over {3 - {x^2}}} = {{\left( {x + \sqrt 3 } \right)\left( {{x^2} - x\sqrt 3 + 3} \right)} \over {\left( {x + \sqrt 3 } \right)\left( {\sqrt 3 - x} \right)}} = {{{x^2} - x\sqrt 3 + 3} \over {\sqrt 3 - x}}\) với \(\,x \ne - \sqrt 3 \) Do đó : \(\mathop {\lim }\limits_{x \to - \sqrt 3 } {{{x^3} + 3\sqrt 3 } \over {3 - {x^2}}} =\mathop {\lim }\limits_{x \to - \sqrt 3 } {{{x^2} - x\sqrt 3 + 3} \over {\sqrt 3 - x}}= {9 \over {2\sqrt 3 }} = {{3\sqrt 3 } \over 2}\) b. \(\eqalign{ & \mathop {\lim }\limits_{x \to 4} {{\sqrt x - 2} \over {{x^2} - 4x}} = \mathop {\lim }\limits_{x \to 4} {{\sqrt x - 2} \over {x\left( {x - 4} \right)}} \cr & = \mathop {\lim }\limits_{x \to 4} {1 \over {x\left( {\sqrt x + 2} \right)}} = {1 \over {16}} \cr} \) c. \(\eqalign{ & \mathop {\lim }\limits_{x \to {1^ + }} {{\sqrt {x - 1} } \over {{x^2} - x}} = \mathop {\lim }\limits_{x \to {1^ + }} {{\sqrt {x - 1} } \over {x\left( {x - 1} \right)}} \cr & = \mathop {\lim }\limits_{x \to {1^ + }} {1 \over {x\sqrt {x - 1} }} = + \infty \cr} \) d. \(\eqalign{ & \mathop {\lim }\limits_{x \to 0} {{\sqrt {{x^2} + x + 1} - 1} \over {3x}} = \mathop {\lim }\limits_{x \to 0} {{{x^2} + x + 1 - 1} \over {3x(\sqrt {{x^2} + x + 1} + 1)}} \cr & = {1 \over 3}\mathop {\lim }\limits_{x \to 0} {{x + 1} \over {\sqrt {{x^2} + x + 1} + 1}} = {1 \over 6} \cr} \) Câu 44 trang 167 SGK Đại số và Giải tích 11 Nâng cao. Tìm các giới hạn sau : a. \(\mathop {\lim }\limits_{x \to - \infty } x\sqrt {{{2{x^3} + x} \over {{x^5} - {x^2} + 3}}} \) b. \(\mathop {\lim }\limits_{x \to - \infty } {{\left| x \right| + \sqrt {{x^2} + x} } \over {x + 10}}\) c. \(\mathop {\lim }\limits_{x \to + \infty } {{\sqrt {2{x^4} + {x^2} - 1} } \over {1 - 2x}}\) d. \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {2{x^2} + 1} + x} \right)\) Giải: a. Với \(x < 0\), ta có : \(\eqalign{ & x\sqrt {{{2{x^3} + x} \over {{x^5} - {x^2} + 3}}} = - \left| x \right|\sqrt {{{2{x^3} + x} \over {{x^5} - {x^2} + 3}}} \cr & = - \sqrt {{{{x^2}\left( {2{x^3} + x} \right)} \over {{x^5} - {x^2} + 3}}} = - \sqrt {{{2 + {1 \over {{x^2}}}} \over {1 - {1 \over {{x^3}}} + {1 \over {{x^5}}}}}} \cr} \) Do đó : \(\mathop {\lim }\limits_{x \to - \infty } x\sqrt {{{2{x^3} + x} \over {{x^5} - {x^2} + 3}}} = - \sqrt 2 \) b. \(\eqalign{ & \mathop {\lim }\limits_{x \to - \infty } {{\left| x \right|+\sqrt {{x^2} + x} } \over {x + 10}} = \mathop {\lim }\limits_{x \to - \infty } {{\left| x \right| + \left| x \right|\sqrt {1 + {1 \over x}} } \over {x + 10}} \cr & = \mathop {\lim }\limits_{x \to - \infty } {{ - x - x\sqrt {1 + {1 \over x}} } \over {x + 10}} = \mathop {\lim }\limits_{x \to - \infty } {{ - 1 - \sqrt {1 + {1 \over x}} } \over {1 + {{10} \over x}}} \cr &= - 2 \cr} \) c. \(\eqalign{ & \mathop {\lim }\limits_{x \to + \infty } {{\sqrt {2{x^4} + {x^2} - 1} } \over {1 - 2x}} = \mathop {\lim }\limits_{x \to + \infty } {{{x^2}\sqrt {2 + {1 \over {{x^2}}} - {1 \over {{x^4}}}} } \over {x\left( {{1 \over x} - 2} \right)}} \cr & = \mathop {\lim }\limits_{x \to + \infty } x{{\sqrt {2 + {1 \over {{x^2}}} - {1 \over {{x^4}}}} } \over {{1 \over x} - 2}} = - \infty \cr & \text{vì}\,\mathop {\lim }\limits_{x \to + \infty } x = + \infty \,\text{và}\,\mathop {\lim }\limits_{x \to + \infty } {{\sqrt {2 + {1 \over {{x^2}}} - {1 \over {{x^4}}}} } \over {{1 \over x} - 2}} = - {{\sqrt 2 } \over 2} < 0 \cr} \) d. \(\eqalign{ & \mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {2{x^2} + 1} + x} \right) = \mathop {\lim }\limits_{x \to - \infty } {{2{x^2} + x - {x^2}} \over {\sqrt {2{x^2} + x} - x}} \cr & = \mathop {\lim }\limits_{x \to - \infty } {{x\left( {x + 1} \right)} \over { - x\left( {\sqrt {2 + {1 \over x}} + 1} \right)}} \cr &= \mathop {\lim }\limits_{x \to - \infty } - {{x + 1} \over {\sqrt {2 + {1 \over x} + 1} }} = + \infty \cr & \text{vì }\,\mathop {\lim }\limits_{x \to - \infty } \left( { - x - 1} \right) = + \infty \cr} \) Câu 45 trang 167 SGK Đại số và Giải tích 11 Nâng cao. Tìm các giới hạn sau : a. \(\mathop {\lim }\limits_{x \to {0^ + }} {{\sqrt {{x^2} + x} - \sqrt x } \over {{x^2}}}\) b. \(\mathop {\lim }\limits_{x \to {1^ - }} x{{\sqrt {1 - x} } \over {2\sqrt {1 - x} + 1 - x}}\) c. \(\mathop {\lim }\limits_{x \to {3^ - }} {{3 - x} \over {\sqrt {27 - {x^3}} }}\) d. \(\mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {{x^3} - 8} } \over {{x^2} - 2x}}\) Giải: a. \(\eqalign{ & \mathop {\lim }\limits_{x \to {0^ + }} {{\sqrt {{x^2} + x} - \sqrt x } \over {{x^2}}} = \mathop {\lim }\limits_{x \to {0^ + }} {x^2 \over {{x^2}\left( {\sqrt {{x^2} + x} + \sqrt x } \right)}} \cr & = \mathop {\lim }\limits_{x \to {0^ + }} {1 \over {\left( {\sqrt {{x^2} + x} + \sqrt x } \right)}} = + \infty \cr} \) b. \(\mathop {\lim }\limits_{x \to {1^ - }} x{{\sqrt {1 - x} } \over {2\sqrt {1 - x} + 1 - x}} = \mathop {\lim }\limits_{x \to {1^ - }} {x \over {2 + \sqrt {1 - x} }} = {1 \over 2}\) c. \(\eqalign{ & \mathop {\lim }\limits_{x \to {3^ - }} {{3 - x} \over {\sqrt {27 - {x^3}} }} = \mathop {\lim }\limits_{x \to {3^ - }} {{{{\left( {\sqrt {3 - x} } \right)}^2}} \over {\sqrt {\left( {3 - x} \right)\left( {{x^2} + 3x + 9} \right)} }} \cr & = \mathop {\lim }\limits_{x \to {3^ - }} {{\sqrt {3 - x} } \over {\sqrt {{x^2} + 3x + 9} }} = 0 \cr} \) d. \(\eqalign{ & \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {{x^3} - 8} } \over {{x^2} - 2x}} = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)} } \over {x\left( {x - 2} \right)}} \cr & = \mathop {\lim }\limits_{x \to {2^ + }} {1 \over x}\sqrt {{{{x^2} + 2x + 4} \over {x - 2}}} = + \infty \cr} \) Vì \(\eqalign{ & \mathop {\lim }\limits_{x \to {2^ + }} \sqrt {{x^2} + 2x + 4} = 2\sqrt 3 \cr & \mathop {\lim }\limits_{x \to {2^ + }} x\sqrt {x - 2} = 0;\,x\sqrt {x - 2} > 0\,\forall x > 2 \cr} \)
