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a) \(\lim \left[ {f\left( {{x_n}} \right) + g\left( {{x_n}} \right)} \right] = \lim \left( {2{x_n} + \frac{{{x_n}}}{{{x_n} + 1}}} \right) = 2\lim {x_n} + \lim \frac{{{x_n}}}{{{x_n} + 1}} = 2.1 + \frac{1}{{1 + 1}} = \frac{5}{2}\)
b) Vì \(\lim \left[ {f\left( {{x_n}} \right) + g\left( {{x_n}} \right)} \right] = \frac{5}{2}\) nên \(\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) + g\left( x \right)} \right] = \frac{5}{2}\) (1).
Ta có: \(\lim {\rm{ }}f\left( {{x_n}} \right) = \lim 2{x_n} = 2\lim {x_n} = 2.1 = 2 \Rightarrow \mathop {\lim }\limits_{x \to 1} {\rm{ }}f\left( x \right) = 2\)
\(\lim g\left( {{x_n}} \right) = \lim \frac{{{x_n}}}{{{x_n} + 1}} = \lim \frac{{{x_n}}}{{{x_n} + 1}} = \frac{1}{{1 + 1}} = \frac{1}{2} \Rightarrow \mathop {\lim }\limits_{x \to 1} {\rm{ }}g\left( x \right) = \frac{1}{2}\)
Vậy \(\mathop {\lim }\limits_{x \to 1} {\rm{ }}f\left( x \right) + \mathop {\lim }\limits_{x \to 1} g\left( x \right) = 2 + \frac{1}{2} = \frac{5}{2}\) (2).
Từ (1) và (2) suy ra \(\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) + g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} {\rm{ }}f\left( x \right) + \mathop {\lim }\limits_{x \to 1} g\left( x \right)\)
\(\lim\limits_{x\rightarrow x_0}f\left(x\right)=+\infty\)
Ta có xn luôn dương
Ta có \(2x_n+1=\) \(2\times\dfrac{\left(2+cos\alpha\right)x_n+cos^2\alpha}{\left(2-2cos2\alpha\right)x_n+2-cos2\alpha}+1=\)
\(=\dfrac{6x_n+2cos^2\alpha+2-cos2\alpha}{\left(2-2cos2\alpha\right)x_n+2-cos2\alpha}\)
\(=\dfrac{6x_n+2cos^2\alpha+2sin^2a+1}{\left(2x_n+1\right)\left(1-cos2\alpha\right)+1}\)
\(=\dfrac{3\left(2x_n+1\right)}{2\sin^2\alpha\left(2x_n+1\right)+1}\)
\(\Rightarrow\dfrac{1}{2x_{n+1}+1}=\dfrac{2\sin^2\alpha\left(2x_n+1\right)+1}{3\left(2x_n+1\right)}\)
\(=\dfrac{1}{3}\left(2\sin^2\alpha+\dfrac{1}{2x_n+1}\right)\)
\(\Rightarrow\dfrac{1}{2x_{n+1}+1}-\sin^2\alpha=\dfrac{1}{3}\left(\dfrac{1}{2x_n+1}-\sin^2\alpha\right)\)
\(\Rightarrow\dfrac{1}{2x_{n+1}+1}-\sin^2\alpha=\left(\dfrac{1}{3}\right)^n\left(\dfrac{1}{2x_1+1}-\sin^2\alpha\right)\)
\(=\left(\dfrac{1}{3}\right)^n\left(\dfrac{1}{3}-\sin^2\alpha\right)\)
\(\Rightarrow y_n=\sum\limits^{n-1}_{i=0}\left(\dfrac{1}{3}\right)^i\left(\dfrac{1}{3}-\sin^2\alpha\right)+n\sin^2\alpha\)
\(=\dfrac{1-\left(\dfrac{1}{3}\right)^n}{1-\dfrac{1}{3}}\left(\dfrac{1}{3}-\sin^2\alpha\right)+n\sin^2\alpha\)
a) Ta có: \({u_{n + 1}} = \frac{{2\left( {n + 1} \right) - 1}}{{\left( {n + 1} \right) + 1}} = \frac{{2n + 2 - 1}}{{n + 1 + 1}} = \frac{{2n + 1}}{{n + 2}}\)
Xét hiệu:
\(\begin{array}{l}{u_{n + 1}} - {u_n} = \frac{{2n + 1}}{{n + 2}} - \frac{{2n - 1}}{{n + 1}} = \frac{{\left( {2n + 1} \right)\left( {n + 1} \right) - \left( {2n - 1} \right)\left( {n + 2} \right)}}{{\left( {n + 2} \right)\left( {n + 1} \right)}}\\ = \frac{{\left( {2{n^2} + n + 2n + 1} \right) - \left( {2{n^2} - n + 4n - 2} \right)}}{{\left( {n + 2} \right)\left( {n + 1} \right)}}\\ = \frac{{2{n^2} + n + 2n + 1 - 2{n^2} + n - 4n + 2}}{{\left( {n + 2} \right)\left( {n + 1} \right)}} = \frac{3}{{\left( {n + 2} \right)\left( {n + 1} \right)}} > 0,\forall n \in {\mathbb{N}^*}\end{array}\)
Vậy \({u_{n + 1}} - {u_n} > 0 \Leftrightarrow {u_{n + 1}} > {u_n}\). Vậy dãy số \(\left( {{u_n}} \right)\) là dãy số tăng.
b) Ta có: \({x_{n + 1}} = \frac{{\left( {n + 1} \right) + 2}}{{{4^{n + 1}}}} = \frac{{n + 1 + 2}}{{{{4.4}^n}}} = \frac{{n + 3}}{{{{4.4}^n}}}\)
Xét hiệu:
\({x_{n + 1}} - {x_n} = \frac{{n + 3}}{{{{4.4}^n}}} - \frac{{n + 2}}{{{4^n}}} = \frac{{n + 3 - 4\left( {n + 2} \right)}}{{{{4.4}^n}}} = \frac{{n + 3 - 4n - 8}}{{{{4.4}^n}}} = \frac{{ - 3n - 5}}{{{{4.4}^n}}} < 0,\forall n \in {\mathbb{N}^*}\)
Vậy \({x_{n + 1}} - {x_n} < 0 \Leftrightarrow {x_{n + 1}} < {x_n}\). Vậy dãy số \(\left( {{x_n}} \right)\) là dãy số giảm.
c) Ta có: \({t_1} = {\left( { - 1} \right)^1}{.1^2} = - 1;{t_2} = {\left( { - 1} \right)^2}{.2^2} = 4;{t_3} = {\left( { - 1} \right)^3}{.3^2} = - 9\), suy ra \({t_1} < {t_2},{t_2} > {t_3}\). Vậy \(\left( {{t_n}} \right)\) là dãy số không tăng không giảm.
a) Ta có: \({u_{n + 1}} = \frac{{2\left( {n + 1} \right) - 1}}{{\left( {n + 1} \right) + 1}} = \frac{{2n + 2 - 1}}{{n + 1 + 1}} = \frac{{2n + 1}}{{n + 2}}\)
Xét hiệu:
\(\begin{array}{l}{u_{n + 1}} - {u_n} = \frac{{2n + 1}}{{n + 2}} - \frac{{2n - 1}}{{n + 1}} = \frac{{\left( {2n + 1} \right)\left( {n + 1} \right) - \left( {2n - 1} \right)\left( {n + 2} \right)}}{{\left( {n + 2} \right)\left( {n + 1} \right)}}\\ = \frac{{\left( {2{n^2} + n + 2n + 1} \right) - \left( {2{n^2} - n + 4n - 2} \right)}}{{\left( {n + 2} \right)\left( {n + 1} \right)}}\\ = \frac{{2{n^2} + n + 2n + 1 - 2{n^2} + n - 4n + 2}}{{\left( {n + 2} \right)\left( {n + 1} \right)}} = \frac{3}{{\left( {n + 2} \right)\left( {n + 1} \right)}} > 0,\forall n \in {\mathbb{N}^*}\end{array}\)
Vậy \({u_{n + 1}} - {u_n} > 0 \Leftrightarrow {u_{n + 1}} > {u_n}\). Vậy dãy số \(\left( {{u_n}} \right)\) là dãy số tăng.
b) Ta có: \({x_{n + 1}} = \frac{{\left( {n + 1} \right) + 2}}{{{4^{n + 1}}}} = \frac{{n + 1 + 2}}{{{{4.4}^n}}} = \frac{{n + 3}}{{{{4.4}^n}}}\)
Xét hiệu:
\({x_{n + 1}} - {x_n} = \frac{{n + 3}}{{{{4.4}^n}}} - \frac{{n + 2}}{{{4^n}}} = \frac{{n + 3 - 4\left( {n + 2} \right)}}{{{{4.4}^n}}} = \frac{{n + 3 - 4n - 8}}{{{{4.4}^n}}} = \frac{{ - 3n - 5}}{{{{4.4}^n}}} < 0,\forall n \in {\mathbb{N}^*}\)
Vậy \({x_{n + 1}} - {x_n} < 0 \Leftrightarrow {x_{n + 1}} < {x_n}\). Vậy dãy số \(\left( {{x_n}} \right)\) là dãy số giảm.
c) Ta có: \({t_1} = {\left( { - 1} \right)^1}{.1^2} = - 1;{t_2} = {\left( { - 1} \right)^2}{.2^2} = 4;{t_3} = {\left( { - 1} \right)^3}{.3^2} = - 9\), suy ra \({t_1} < {t_2},{t_2} > {t_3}\). Vậy \(\left( {{t_n}} \right)\) là dãy số không tăng không giảm.
Tham khảo:
a,
\(\lim f\left( {{x_n}} \right) = \lim \left( {2.\frac{{n + 1}}{n}} \right) = \lim 2.\lim \left( {1 + \frac{1}{n}} \right) = 2.\left( {1 + 0} \right) = 2\)
b) Lấy dãy số bất kì \(\left( {{x_n}} \right),{x_n} \to 1\) ta có \(f\left( {{x_n}} \right) = 2{x_n}.\)
\(\lim f\left( {{x_n}} \right) = \lim \left( {2{x_n}} \right) = \lim 2.\lim {x_n} = 2.1 = 2\)