22 câu hỏi
\(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) + 3} \right] = 6\).
\(\mathop {\lim }\limits_{x \to 2} \left( {f\left( x \right) - 1} \right) = 2\).
\(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) - 2x} \right] = - 1\).
\(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) - {x^2}} \right] = 1\).
D
\(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) + 3} \right] = \mathop {\lim }\limits_{x \to 2} f\left( x \right) + 3 = 3 + 3 = 6\).
\(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) - 2x} \right] = \mathop {\lim }\limits_{x \to 2} f\left( x \right) - \mathop {\lim }\limits_{x \to 2} 2x = 3 - 4 = - 1\).
\(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) - {x^2}} \right] = \mathop {\lim }\limits_{x \to 2} f\left( x \right) - \mathop {\lim }\limits_{x \to 2} {x^2} = 3 - 4 = - 1\).
\(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right) - 1} \right] = \mathop {\lim }\limits_{x \to 2} f\left( x \right) - 1 = 3 - 1 = 2\).
\(5\).
\(2\).
\( - 6\).
\(3\).
C
Ta có \(\mathop {\lim }\limits_{x \to {x_0}} \left[ {3f\left( x \right) - 4g\left( x \right)} \right]\)\( = \mathop {\lim }\limits_{x \to {x_0}} 3f\left( x \right) - \mathop {\lim }\limits_{x \to {x_0}} 4g\left( x \right)\)\( = 3\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) - 4\mathop {\lim }\limits_{x \to {x_0}} g\left( x \right)\)\( = - 6\).
\[2\].
\[1\].
\[ + \infty \].
\[0\].
D
Ta có: \[\mathop {\lim }\limits_{x \to 1} \left( {2{x^2} - 3x + 1} \right) = 0\].
0.
\[\frac{{ - 1}}{7}\].
−7.
+∞.
B
\(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 5x + 6}}{{{x^3} - {x^2} - x - 2}}\)\( = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + x + 1} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{x - 3}}{{{x^2} + x + 1}} = - \frac{1}{7}\).
−∞.
1.
+∞.
−1.
B
\(\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} + 2} - 2}}{{x - 2}}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\sqrt {1 + \frac{2}{{{x^2}}}} - 2}}{{x - 2}}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {1 + \frac{2}{{{x^2}}}} - \frac{2}{x}}}{{1 - \frac{2}{x}}} = 1\).
\(\frac{1}{3}\).
\(\frac{1}{2}\).
\( - \frac{1}{3}\).
\( - \frac{1}{2}\).
C
Ta có \[\mathop {\lim }\limits_{x \to - \infty } \frac{{1 - x}}{{3x + 2}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{1}{x} - 1}}{{3 + \frac{2}{x}}} = - \frac{1}{3}\].
\[ - \frac{1}{4}\].
\[1\].
\[0\].
\[\frac{1}{4}\].
A
Ta có \(\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} + 3x + 5} }}{{4x - 1}}\)\( = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {1 + \frac{3}{x} + \frac{5}{{{x^2}}}} }}{{4 - \frac{1}{x}}} = - \frac{1}{4}\).
Học thông minh hơn mỗi ngày
Xem toàn bộ 22 câu hỏi và nhiều hơn nữa với tài khoản miễn phí
\( + \infty .\)
\( - \infty .\)
\(\frac{2}{3}.\)
\(\frac{1}{3}.\)
B
Ta có \(\mathop {\lim }\limits_{x \to {1^ + }} \left( { - 2x + 1} \right) = - 1 < 0\), \(\mathop {\lim }\limits_{x \to {1^ + }} \left( {x - 1} \right) = 0\), \[x - 1 > 0\] khi \[x \to {1^ + }\].
Suy ra \(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{ - 2x + 1}}{{x - 1}} = - \infty \).
\[ + \infty \].
\[\frac{1}{2}\].
\[ - \infty \]
\[ - \frac{1}{2}\].
C
\(\mathop {\lim }\limits_{x \to {1^ - }} \frac{{x + 2}}{{x - 1}} = - \infty \) vì \[\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left( {x + 2} \right) = 3 > 0\\\mathop {\lim }\limits_{x \to 1} \left( {x - 1} \right) = 0\\x - 1 < 0,\forall x < 1\end{array} \right.\].
\[\frac{1}{2}\].
\[ - \frac{1}{2}\].
\[\frac{3}{2}\]
\[ - \frac{3}{2}\].
D
Ta có: \[\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \frac{{\sqrt {3{x^2} + 1} - x}}{{x - 1}} = \frac{{\sqrt 4 + 1}}{{ - 1 - 1}} = - \frac{3}{2}\].
\( + \infty .\)
−1.
0.
1.
A
\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)\)\( = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{{x^2} + 1}}{{1 - x}} = + \infty \).
\(0\).
\( + \infty \).
\( - \infty \).
\( - 4\).
B
Ta có \(\mathop {\lim }\limits_{x \to - \infty } \left( { - 4{x^5} - 3{x^3} + x + 1} \right)\)\( = \mathop {\lim }\limits_{x \to - \infty } {x^5}\left( { - 4 - \frac{3}{{{x^2}}} + \frac{1}{{{x^4}}} + \frac{1}{{{x^5}}}} \right)\)\( = + \infty \).
Vì \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \left( { - 4 - \frac{3}{{{x^2}}} + \frac{1}{{{x^4}}} + \frac{1}{{{x^5}}}} \right) = - 4 < 0\\\mathop {\lim }\limits_{x \to - \infty } {x^5} = - \infty \end{array} \right.\).
a) \(\mathop {\lim }\limits_{x \to 2} \left[ {5f\left( x \right)} \right] = 5\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 5.5 = 25\).
b) \(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right).g\left( x \right)} \right] = + \infty \).
c) \[\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right)}}{{g\left( x \right)}} = 0\].
d) \(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {f\left( x \right) - 1} - 2}}{{f\left( x \right) - 5}} = \mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 5}}{{\left( {\sqrt {f\left( x \right) - 1} + 2} \right)\left( {f\left( x \right) - 5} \right)}}\)\( = \mathop {\lim }\limits_{x \to 2} \frac{1}{{\sqrt {f\left( x \right) - 1} + 2}} = \frac{1}{4}\).
Đáp án: a) Sai; b) Đúng; c) Sai; d) Đúng.
a) \(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \sqrt {{x^2} + 1} = \sqrt 5 \).
b) \(\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {{x^2} - 3x + 1} \right) = 1\).
c) \(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \sqrt {{x^2} + 1} = 1\).
d) Do \(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)\) nên \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = 1\).
Đáp án: a) Sai; b) Sai; c) Đúng; d) Đúng.
a) \(f\left( 8 \right) = \frac{{\sqrt {8 + 1} - 2}}{{8 - 3}} = \frac{1}{5}\).
b) \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {x + 1} - 2}}{{x - 3}} = \frac{1}{3}\).
c) \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \frac{{x - 3}}{{\left( {\sqrt {x + 1} + 2} \right)\left( {x - 3} \right)}} = \mathop {\lim }\limits_{x \to 3} \frac{1}{{\sqrt {x + 1} + 2}} = \frac{1}{4}\).
d) \(\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\sqrt {\frac{1}{x} + \frac{1}{{{x^2}}}} - 2}}{{x - 3}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {\frac{1}{x} + \frac{1}{{{x^2}}}} - \frac{2}{x}}}{{1 - \frac{3}{x}}} = 0\).
\(\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + x + 2} - x} \right)\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{x + 2}}{{\sqrt {{x^2} + x + 2} + x}}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{1 + \frac{2}{x}}}{{\sqrt {1 + \frac{1}{x} + \frac{2}{{{x^2}}}} + 1}} = \frac{1}{2}\).
Do đó 3a + 4b = 2.
Đáp án: a) Sai; b) Đúng; c) Sai; d) Đúng.
a) \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 3x + 2}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 2} \right)\left( {x - 1} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {x - 2} \right) = - 1\).
b) \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 3x + 2}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 2} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{x - 2}}{{x + 1}} = - \frac{1}{2}\).
c) \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{{x^3} - {x^2} + x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 2} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{x - 2}}{{{x^2} + 1}} = - \frac{1}{2} < 0\).
d) Để \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{ax + b}} = 2\) thì ax + b có nghiệm bằng 1 Û a + b = 0 Û b = −a.
Khi đó \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{ax + b}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 2} \right)\left( {x - 1} \right)}}{{a\left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{x - 2}}{a} = - \frac{1}{a} = 2\) \( \Leftrightarrow a = - \frac{1}{2} \Rightarrow b = \frac{1}{2}\).
Suy ra \(a + 3b = - \frac{1}{2} + 3.\frac{1}{2} = 1\).
Đáp án: a) Đúng; b) Sai; c) Sai; d) Đúng.
a) Ta có f(2) = 2a + 2024.
b) \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} - 4}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \left( {x + 2} \right) = 4\).
c) \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {ax + 2024} \right) = 2a + 2024\).
d) Hàm số f(x) có giới hạn khi x → 2 khi và chỉ khi \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)\)
Û 4 = 2a + 2024 Û a = −1010.
Đáp án: a) Sai; b) Đúng; c) Sai; d) Đúng.
Sau t phút bơm nước vào hồ thì lượng nước là 600 + 15t (lít) và lượng muối có được là 30.15t gam.
Nồng độ muối của nước là \(C\left( t \right) = \frac{{30.15t}}{{600 + 15t}} = \frac{{30t}}{{40 + t}}\) (gam/lít).
Khi t dần về dương vô cùng, ta có \(\mathop {\lim }\limits_{t \to + \infty } C\left( t \right) = \mathop {\lim }\limits_{t \to + \infty } \frac{{30t}}{{40 + t}} = \mathop {\lim }\limits_{t \to + \infty } \frac{{30}}{{\frac{{40}}{t} + 1}} = 30\) (gam/lít).
Trả lời: 30.
Ta có \(\mathop {\lim }\limits_{x \to - \infty } \left[ {\left( {{m^2} - 4m + 3} \right){x^4} - x + 2025} \right] = - \infty \)
\( \Leftrightarrow \mathop {\lim }\limits_{x \to - \infty } {x^4}\left[ {\left( {{m^2} - 4m + 3} \right) - \frac{1}{{{x^3}}} + \frac{{2025}}{{{x^4}}}} \right] = - \infty \).
Vì \(\mathop {\lim }\limits_{x \to - \infty } {x^4} = + \infty \) và \(\mathop {\lim }\limits_{x \to - \infty } \left( {{m^2} - 4m + 3 - \frac{1}{{{x^3}}} + \frac{{2025}}{{{x^4}}}} \right) = {m^2} - 4m + 3\).
Để \(\mathop {\lim }\limits_{x \to - \infty } {x^4}\left[ {\left( {{m^2} - 4m + 3} \right) - \frac{1}{{{x^3}}} + \frac{{2025}}{{{x^4}}}} \right] = - \infty \)thì m2 – 4m + 3 < 0 Û 1 < m < 3.
Mà m Î ℤ nên m = 2.
Vậy có 1 giá trị nguyên.
Trả lời: 1.
Ta có \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2} + ax + 2} \right) = a + 3\) và \(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {2{x^2} - x + 3a} \right) = 3a + 1\).
Hàm số có giới hạn khi x → 1 khi và chỉ khi \(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)\) Û a + 3 = 3a + 1 Û a = 1.
Trả lời: 1.
\(\mathop {\lim }\limits_{x \to 3} \frac{{2{x^2} - 5x - 3}}{{\sqrt {5x + 1} - 4}}\)\( = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {2x + 1} \right)\left( {x - 3} \right)\left( {\sqrt {5x + 1} + 4} \right)}}{{5\left( {x - 3} \right)}}\)
\( = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {2x + 1} \right)\left( {\sqrt {5x + 1} + 4} \right)}}{5} = \frac{{56}}{5} = 11,2\).
Trả lời: 11,2.
\(\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x + 1} - 2}}{{x - 3}}\)\( = \mathop {\lim }\limits_{x \to 3} \frac{{x - 3}}{{\left( {x - 3} \right)\left( {\sqrt {x + 1} + 2} \right)}}\)\( = \mathop {\lim }\limits_{x \to 3} \frac{1}{{\sqrt {x + 1} + 2}}\)\( = \frac{1}{{{2^2}}}\).
Suy ra \[a = 1;\,b = 2\].
\(\sqrt a + b + 2018 = 1 + 2 + 2018 = 2021\).
Trả lời: 2021.
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