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monutza
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Pe: 2009-10-27T19:19:53+02:00 In: In:

va rog sa ma ajutati urgent !!!!!!!!!! plsss:(:(

a) x(indice n)=cos n pi/3

b)yn=sin npi/2

c) lim 1/2 la n sin n
(cand n tinde la infinit)

d) lim [i/nla a 2-a -(i/3)la n] cos n
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e) lim 1/n la a 2-a (3sin n- 6 cosn)
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f) lim 1/3 la n (sqrt(n+1) -sqrt n )
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Similare

4 raspunsuri

  1. ali
    maestru (V)

    lim 1/2 la n sin n
    (cand n tinde la infinit)


    Rezolvare:
    Se poate rezolva prin mai multe metode, una din ele ar fi:
    \[ {\lim }\limits_{n \to \infty } \left( {\frac{1}{2}} \right)^n \sin n = 0;\;\;\;pt.ca\;]

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  2. ali
    maestru (V)

    lim 1/n la a 2-a (3sin n- 6 cosn)


    Rezolvare:

        \[ \begin{array}{l} \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\rm{1}}}{{\rm{n}}}} \right)^{\rm{2}} \left( {3\sin {\rm{ }}n - 6{\rm{ }}\cos n} \right) = 3\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\sin n}}{{{\rm{n}}^2 }}} \right) - 6\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\cos n}}{{{\rm{n}}^2 }}} \right) \\ \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\sin n}}{{{\rm{n}}^2 }}} \right) = 0,\;{\rm{folosind criteriul majorarii}}{\rm{.}} \\ \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\cos n}}{{{\rm{n}}^2 }}} \right) = 0,\;{\rm{folosind criteriul majorarii}}{\rm{.}} \\ \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\rm{1}}}{{\rm{n}}}} \right)^{\rm{2}} \left( {3\sin {\rm{ }}n - 6{\rm{ }}\cos n} \right) = 0 \\ \end{array} \]

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  4. ali
    maestru (V)

    a) x(indice n)=cos n pi/3
    b)yn=sin npi/2


    Rezolvare:
    Cele doua siruri nu au limite.se studeaza pe cazuri:
    n=2k
    n=2k+1

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  5. ali
    maestru (V)

    lim 1/3 la n (sqrt(n+1) -sqrt n )


    Rezolvare:

        \[ \begin{array}{l} {\lim }\limits_{n \to \infty } \left( {\frac{{\rm{1}}}{{\rm{3}}}} \right)^{\rm{n}} \left( {\sqrt {{\rm{n + 1}}} - \sqrt {\rm{n}} } \right) = \left( {\frac{{\rm{1}}}{{\rm{3}}}} \right)^{\rm{n}} {\lim }\limits_{n \to \infty } \left( {\sqrt {{\rm{n + 1}}} - \sqrt {\rm{n}} } \right) \\ {\lim }\limits_{n \to \infty } \left( {\sqrt {{\rm{n + 1}}} - \sqrt {\rm{n}} } \right) = {\lim }\limits_{n \to \infty } \frac{{{\rm{n + 1}} - n}}{{\sqrt {{\rm{n + 1}}} + \sqrt {\rm{n}} }} = {\lim }\limits_{n \to \infty } \frac{1}{{\sqrt {{\rm{n + 1}}} + \sqrt {\rm{n}} }} = 0 \\ {\lim }\limits_{n \to \infty } \left( {\frac{{\rm{1}}}{{\rm{3}}}} \right)^{\rm{n}} \left( {\sqrt {{\rm{n + 1}}} - \sqrt {\rm{n}} } \right) = 0 \\ \end{array} \]

    ❗ Am sa te rog ca in viitor sa alegi un titlu potrivit solicitarilor tale.

    • 0
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