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justmegeo
Pe: 2012-11-12T20:20:11+02:00 In: In:

Limita

Buna seara! Ma poate ajuta cineva sa solutionez niste exercitii cu limite ? 🙂
Am incercat sa le rezolv, insa nu stiu daca este corect.

1. Sa se calculeze limitele sirurilor:
a) an= [1/(n^2+1)]*(1+2+3+…+n)
Rezolvarea mea:
lim an=lim 1/(n^2+1)*n(n+1)/2=lim n(n+1)/2(n^2+1)/2(n^2+1)=lim n^2+n/2n^2+2=1/2
b)an=( 1^2+2^2+3^2+…+n^2 )/n(1+2+3+….+n);
Rezolvarea mea:
lim an=lim [n(n+1)(2n+1)/6]/n^2(n+1)/2=lim 2n(n+1)(2n+1)/6*n^2(n+1)=
lim n(2n+1)/3n^2= lim (2n^2+n)/3n^2 =2/3

2.Sa se calculeze limitele sirurilor:
a) (am sa notez cu nv – radical de ordin n)
an=n(nv2 – 1)
lim an = n(n^1/n -1) = ln 2

b)an= (nv3-nv2)/(nv5-nv3)
lim an= lim (3^1/n – 2^1/n) / (5^1/n – 3^1/n) =0

Va rog frumos daca puteti arunca o privire in decursul zilei de astazi. Multumesc foarte mult pt sprijinul si ajutorul acordat.

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1 raspuns

  1. xor_NTG
    maestru (V)

    La 2 a) ai asa:

        \[ {\lim }\limits_{n \to \infty } n\left( {\sqrt[n]{2} - 1} \right) = {\lim }\limits_{n \to \infty } n\left( {2^{\frac{1}{n}} - 1} \right) = {\lim }\limits_{n \to \infty } n\frac{{2^{\frac{1}{n}} - 1}}{{\frac{1}{n}}}\frac{1}{n} = \ln 2 \]

    Observ ca ti-a dat bine, dar ai pus acolo n^{\frac{1}{n}} in loc de 2^{\frac{1}{n}}

    La b) ai nedeterminarea \frac{0}{0}. Eu am rezolvat asa:

        \[ a_n = \frac{{3^{\frac{1}{n}} - 2^{\frac{1}{n}} }}{{5^{\frac{1}{n}} - 3^{\frac{1}{n}} }} = \frac{{2^{\frac{1}{n}} \left( {\left( {\frac{3}{2}} \right)^{\frac{1}{n}} - 1} \right)}}{{3^{\frac{1}{n}} \left( {\left( {\frac{5}{3}} \right)^{\frac{1}{n}} - 1} \right)}} = \frac{{2^{\frac{1}{n}} \left( {\frac{{\left( {\frac{3}{2}} \right)^{\frac{1}{n}} - 1}}{{\frac{1}{n}}}\frac{1}{n}} \right)}}{{3^{\frac{1}{n}} \left( {\frac{{\left( {\frac{5}{3}} \right)^{\frac{1}{n}} - 1}}{{\frac{1}{n}}}\frac{1}{n}} \right)}} = \frac{{2^{\frac{1}{n}} \frac{{\ln 3 - \ln 2}}{n}}}{{3^{\frac{1}{n}} \frac{{\ln 5 - \ln 3}}{n}}} = \left( {\frac{2}{3}} \right)^{\frac{1}{n}} \frac{{\ln 3 - \ln 2}}{n}\frac{n}{{\ln 5 - \ln 3}} = \frac{{\ln 3 - \ln 2}}{{\ln 5 - \ln 3}} \]

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