Tìm những giới hạn sau:
a) (mathop {lim }limits_{x to 0} left( {frac{1}{x} + frac{1}{{{x^2}}}} right))
b) (mathop {lim }limits_{x to - 2} frac{{{x^3} + 8}}{{x + 2}})
c) (mathop {lim }limits_{x to 9} frac{{3 - sqrt x }}{{9 - x}})
d) (mathop {lim }limits_{x to 0} frac{{2 - sqrt {4 - x} }}{x})
e) (mathop {lim }limits_{x to + infty } frac{{{x^4} - {x^3} + 11}}{{2x - 7}})
f) (mathop {lim }limits_{x to - infty } frac{{sqrt {{x^4} + 4} }}{{x + 4}})
Hướng dẫn giải:
Câu a:
Câu b:
(mathop {lim }limits_{x to - 2} frac{{{x^3} + 8}}{{x + 2}} = mathop {lim }limits_{x to - 2} frac{{left( {x + 2} right)left( {{x^2} - 2x + 4} right)}}{{x + 2}} = mathop {lim }limits_{x to - 2} left( {{x^2} - 2x + 4} right) = 12)
Câu c:
(mathop {lim }limits_{x to 9} frac{{3 - sqrt x }}{{9 - x}} = mathop {lim }limits_{x to 9} frac{1}{{3 + sqrt x }} = frac{1}{6})
Câu d:
(mathop {lim }limits_{x to 0} frac{{2 - sqrt {4 - x} }}{x} = mathop {lim }limits_{x to 0} frac{{4 - left( {4 - x} right)}}{{xleft( {2 + sqrt {4 - x} } right)}} = mathop {lim }limits_{x to 0} frac{1}{{2 + sqrt {4 - x} }} = frac{1}{4})
Câu e:
(mathop {lim }limits_{x to + infty } frac{{{x^4} - {x^3} + 11}}{{2x - 7}} = mathop {lim }limits_{x to + infty } frac{{{x^3} - {x^2} + frac{{11}}{x}}}{{2 - frac{7}{x}}} = + infty )
Câu f:
Với x < 0, ta sở hữu: (frac{{sqrt {{x^4} + 4} }}{{x + 4}} = frac{{{x^2}sqrt {1 + frac{4}{{{x^4}}}} }}{{x + 4}} = frac{{xsqrt {1 + frac{4}{{{x^4}}}} }}{{1 + frac{4}{x}}})
Tìm những giới hạn sau:
a) (mathop {lim }limits_{x to - sqrt 3 } frac{{{x^3} + 3sqrt 3 }}{{3 - {x^2}}})
b) (mathop {lim }limits_{x to 4} frac{{sqrt x - 2}}{{{x^2} - 4x}})
c) (mathop {lim }limits_{x to {1^ + }} frac{{sqrt {x - 1} }}{{{x^2} - x}})
d) (mathop {lim }limits_{x to 0} frac{{sqrt {{x^2} + x + 1} - 1}}{{3x}})
Hướng dẫn giải:
Câu a:
Ta sở hữu (frac{{{x^3} + 3sqrt 3 }}{{3 - {x^2}}} = frac{{left( {x + sqrt 3 } right)left( {{x^2} - xsqrt 3 + 3} right)}}{{left( {x + sqrt 3 } right)left( {sqrt 3 - x} right)}} = frac{{{x^2} - xsqrt 3 + 3}}{{sqrt 3 - x}})
Do đó (mathop {lim }limits_{x to - sqrt 3 } frac{{{x^3} + 3sqrt 3 }}{{3 - {x^2}}} =mathop {lim }limits_{x to - sqrt 3 } frac{{{x^2} - xsqrt 3 + 3}}{{sqrt 3 - x}} = frac{9}{{2sqrt 3 }} = frac{{3sqrt 3 }}{2})
Câu b:
(mathop {lim }limits_{x to 4} frac{{sqrt x - 2}}{{{x^2} - 4x}} = mathop {lim }limits_{x to 4} frac{{sqrt x - 2}}{{xleft( {x - 4} right)}} = mathop {lim }limits_{x to 4} frac{1}{{xleft( {sqrt x + 2} right)}} = frac{1}{{16}})
Câu c:
(mathop {lim }limits_{x to {1^ + }} frac{{sqrt {x - 1} }}{{{x^2} - x}} = mathop {lim }limits_{x to {1^ + }} frac{{sqrt {x - 1} }}{{xleft( {x - 1} right)}} = mathop {lim }limits_{x to {1^ + }} frac{1}{{xsqrt {x - 1} }} = + infty )
Câu d:
(mathop {lim }limits_{x to 0} frac{{sqrt {{x^2} + x + 1} - 1}}{{3x}} = mathop {lim }limits_{x to 0} frac{{{x^2} + x + 1 - 1}}{{3xleft( {sqrt {{x^2} + x + 1} + 1} right)}} = frac{1}{3}mathop {lim }limits_{x to 0} frac{{x + 1}}{{sqrt {{x^2} + x + 1} + 1}} = frac{1}{6})
Tìm những giới hạn sau:
a) (mathop {lim }limits_{x to - infty } xsqrt {frac{{2{x^3} + x}}{{{x^5} - {x^2} + 3}}} )
b) (mathop {lim }limits_{x to - infty } frac{{left| x right| + sqrt {{x^2} + x} }}{{x + 10}})
c) (mathop {lim }limits_{x to + infty } frac{{sqrt {2{x^4} + {x^2} - 1} }}{{1 - 2x}})
d) (mathop {lim }limits_{x to - infty } left( {sqrt {2{x^2} + 1} + x} right))
Hướng dẫn giải:
Câu a:
Với x < 0, ta sở hữu:
(begin{array}{l}
xsqrt {frac{{2{x^3} + x}}{{{x^5} - {x^2} + 3}}} = - left| x right|sqrt {frac{{2{x^3} + x}}{{{x^5} - {x^2} + 3}}}
= - sqrt {frac{{{x^2}left( {2{x^3} + x} right)}}{{{x^5} - {x^2} + 3}}} = - sqrt {frac{{2 + frac{1}{{{x^2}}}}}{{1 - frac{1}{{{x^3}}} + frac{1}{{{x^5}}}}}}
end{array})
Do đó (mathop {lim }limits_{x to - infty } xsqrt {frac{{2{x^3} + x}}{{{x^5} - {x^2} + 3}}} = - sqrt 2 )
Câu b:
(begin{array}{l}
mathop {lim }limits_{x to - infty } frac{{left| x right| + sqrt {{x^2} + x} }}{{x + 10}} = mathop {lim }limits_{x to - infty } frac{{left| x right| + left| x right|sqrt {1 + frac{1}{x}} }}{{x + 10}}
= mathop {lim }limits_{x to - infty } frac{{ - x - xsqrt {1 + frac{1}{x}} }}{{x + 10}} = mathop {lim }limits_{x to - infty } frac{{ - 1 - sqrt {1 + frac{1}{x}} }}{{1 + frac{{10}}{x}}} = - 2
end{array})
Câu c:
(mathop {lim }limits_{x to + infty } frac{{sqrt {2{x^4} + {x^2} - 1} }}{{1 - 2x}} = mathop {lim }limits_{x to + infty } frac{{{x^2}sqrt {2 + frac{1}{{{x^2}}} - frac{1}{{{x^4}}}} }}{{xleft( {frac{1}{x} - 2} right)}} = mathop {lim }limits_{x to + infty } x.frac{{sqrt {2 + frac{1}{{{x^2}}} - frac{1}{{{x^4}}}} }}{{frac{1}{x} - 2}} = - infty )
(vì (mathop {lim }limits_{x to + infty } x = + infty ,mathop {lim }limits_{x to + infty } frac{{sqrt {2 + frac{1}{{{x^2}}} - frac{1}{{{x^4}}}} }}{{frac{1}{x} - 2}} = - frac{{sqrt 2 }}{2} < 0))
Câu d:
(begin{array}{l}
mathop {lim }limits_{x to - infty } left( {sqrt {2{x^2} + 1} + x} right) = mathop {lim }limits_{x to - infty } frac{{2{x^2} + x - {x^2}}}{{sqrt {2{x^2} + x} - x}}
= mathop {lim }limits_{x to - infty } frac{{xleft( {x + 1} right)}}{{ - xleft( {sqrt {2 + frac{1}{x}} + 1} right)}} = mathop {lim }limits_{x to - infty } - frac{{x + 1}}{{sqrt {2 + frac{1}{x} + 1} }} = + infty
end{array})
(vì (mathop {lim }limits_{x to - infty } left( { - x - 1} right) = + infty ))
Tìm những giới hạn sau:
a) (mathop {lim }limits_{x to {0^ + }} frac{{sqrt {{x^2} + x} - sqrt x }}{{{x^2}}})
b) (mathop {lim }limits_{x to {1^ - }} x.frac{{sqrt {1 - x} }}{{2sqrt {1 - x} + 1 - x}})
c) (mathop {lim }limits_{x to {3^ - }} frac{{3 - x}}{{sqrt {27 - {x^3}} }})
d) (mathop {lim }limits_{x to {2^ + }} frac{{sqrt {{x^3} - 8} }}{{{x^2} - 2x}})
Hướng dẫn giải:
Câu a:
(mathop {lim }limits_{x to {0^ + }} frac{{sqrt {{x^2} + x} - sqrt x }}{{{x^2}}} = mathop {lim }limits_{x to {0^ + }} frac{{{x^2}}}{{{x^2}left( {sqrt {{x^2} + x} + sqrt x } right)}} = mathop {lim }limits_{x to {0^ + }} frac{1}{{sqrt {{x^2} + x} + sqrt x }} = + infty )
Câu b:
(mathop {lim }limits_{x to {1^ - }} x.frac{{sqrt {1 - x} }}{{2sqrt {1 - x} + 1 - x}} = mathop {lim }limits_{x to {1^ - }} frac{x}{{2 + sqrt {1 - x} }} = frac{1}{2})
Câu c:
(begin{array}{l}
mathop {lim }limits_{x to {3^ - }} frac{{3 - x}}{{sqrt {27 - {x^3}} }} = mathop {lim }limits_{x to {3^ - }} frac{{{{left( {sqrt {3 - x} } right)}^2}}}{{sqrt {left( {3 - x} right)left( {{x^2} + 3x + 9} right)} }}
= mathop {lim }limits_{x to {3^ - }} frac{{sqrt {3 - x} }}{{sqrt {{x^2} + 3x + 9} }} = 0
end{array})
Câu d:
(mathop {lim }limits_{x to {2^ + }} frac{{sqrt {{x^3} - 8} }}{{{x^2} - 2x}} = mathop {lim }limits_{x to {2^ + }} frac{{sqrt {left( {x - 2} right)left( {{x^2} + 2x + 4} right)} }}{{xleft( {x - 2} right)}} = mathop {lim }limits_{x to {2^ + }} frac{1}{x}sqrt {frac{{{x^2} + 2x + 4}}{{x - 2}}} = + infty )
Trên đây là nội dung chi tiết Giải bài tập tăng Toán 11 Chương 4 Tập dượt (trang 167) với hướng dẫn giải chi tiết, rõ ràng, trình bày khoa học. Hoc247 kỳ vọng đây sẽ là tài liệu hữu ích giúp những bạn học trò lớp 11 học tập thật tốt.