Câu hỏi:

16/11/2025 136 Lưu

Cho hai dãy số \(\left( {{u_n}} \right)\) và \(\left( {{v_n}} \right)\) với \({u_n} = \frac{{2{n^2} - 4n + 7}}{{8{n^2} + 3n + 10}}\), \({v_n} = \frac{{\sqrt {4{n^2} + 5} }}{{8n}}\).

\(\lim {u_n} = 7\).

\(\lim \left( {{v_n} - \frac{1}{4}} \right) = 0\).

\(\lim \left( {2{u_n} - 4{v_n}} \right) = 0\).

\(\lim \frac{{{u_n}}}{{2{v_n}}} = \frac{1}{2}\).

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Trả lời:

verified Giải bởi Vietjack

a) \(\lim {u_n} = \lim \frac{{2{n^2} - 4n + 7}}{{8{n^2} + 3n + 10}}\)\( = \lim \frac{{2 - \frac{4}{n} + \frac{7}{{{n^2}}}}}{{8 + \frac{3}{n} + \frac{{10}}{{{n^2}}}}} = \frac{1}{4}\).

b) \(\lim {v_n} = \lim \frac{{\sqrt {4{n^2} + 5} }}{{8n}} = \lim \frac{{\sqrt {4 + \frac{5}{{{n^2}}}} }}{8} = \frac{1}{4}\). Suy ra \(\lim \left( {{v_n} - \frac{1}{4}} \right) = 0\).

c) \(\lim \left( {2{u_n} - 4{v_n}} \right) = 2 \cdot \frac{1}{4} - 4 \cdot \frac{1}{4} = - \frac{1}{2}\).

d) \(\lim \frac{{{u_n}}}{{2{v_n}}} = \frac{1}{4}:\left( {2 \cdot \frac{1}{4}} \right) = \frac{1}{2}\).

Đáp án: a) Sai; b) Đúng; c) Sai; d) Đúng.

CÂU HỎI HOT CÙNG CHỦ ĐỀ

Lời giải

Vì \(\mathop {\lim }\limits_{x \to 2} \frac{{a{x^2} + bx - 2}}{{x - 2}} = 5\) và \(\mathop {\lim }\limits_{x \to 2} \left( {x - 2} \right) = 0\) nên \(\mathop {\lim }\limits_{x \to 2} \left( {a{x^2} + bx - 2} \right) = 0\) hay \(4a + 2b - 2 = 0 \Leftrightarrow b = 1 - 2a\).

Khi đó \(\mathop {\lim }\limits_{x \to 2} \frac{{a{x^2} + bx - 2}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{a{x^2} + \left( {1 - 2a} \right)x - 2}}{{x - 2}}\)\( = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {a{x^2} - 2ax} \right) + \left( {x - 2} \right)}}{{x - 2}}\)\( = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {ax + 1} \right)}}{{x - 2}}\)

\( = \mathop {\lim }\limits_{x \to 2} \left( {ax + 1} \right) = 2a + 1 = 5 \Rightarrow a = 2 \Rightarrow b = - 3\).

Vậy \(S = - 4\).

Trả lời: −4.

Lời giải

Ta có \(\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ - }} \frac{{9 - {x^2}}}{{x - 3}} = \mathop {\lim }\limits_{x \to {3^ - }} \frac{{ - \left( {x - 3} \right)\left( {x + 3} \right)}}{{x - 3}} = \mathop {\lim }\limits_{x \to {3^ - }} \left( { - x - 3} \right) = - 6\).

\(\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right)\)\( = \mathop {\lim }\limits_{x \to {3^ + }} \left( {1 - x} \right) = - 2\).

Suy ra \(a = - 6;b = - 2\). Vậy \({a^2} + {b^2} = 40\).

Trả lời: 40.