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Penguin200
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Pe: 2018-07-23T13:40:07+03:00 In: In:

Limite Utc

Buna!Ajutati-ma, va rog la aceste 3 exercitii:

1) \lim_{n->inf} \frac{a+\sqrt{a}+\sqrt[3]{a}+...+\sqrt[n]{a}-n}{ln n}

2)\lim_{n->inf} \frac{5^{n}*n!}{2^{n}*n^{n}}

3)Fie sirul (xn), n>=0.

xn=cos(pi\sqrt{4n^{2}+n+1}).Care este limita sirului?

Multumesc anticipat!

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1 raspuns

  1. PhantomR
    expert (VI)

    Cu Cesaro-Stolz, prima limita devine \lim \frac{\sqrt[n+1]{a}-1}{\ln (1+\frac{1}{n})}=\lim \frac{a^{\frac{1}{n+1}}-1}{\frac{1}{n+1}} \cdot \frac{\frac{1}{n+1}}{\ln (1+\frac{1}{n})}=\lim \frac{a^{\frac{1}{n+1}}-1}{\frac{1}{n+1}} \cdot \lim\frac{1}{\ln [(1+\frac{1}{n})^{n+1}]}\ln a \cdot \frac{1}{\ln e} = \ln a.

    La problema 2, folosind criteriul raportului pentru siruri (), limita raportului este L=\lim \frac{5(n+1)}{2(n+1)^{n+1}}\cdot n^n=\frac{5}{2}\lim \frac{1}{\frac{(n+1)^n}{n^n}}=\frac{5}{2}\lim \frac{1}{(1+\frac{1}{n})^n}=\frac{5}{2e}. Cum e>2.7, obtinem 0<L<1, deci limita sirului initial este zero.

    Pentru a treia, avem x_n=\cos(\pi \sqrt{4n^2+n+1}-2n\pi)=\cos \pi(\sqrt{4n^2+n+1}-2n)=\cos \pi \frac{n+1}{\sqrt{4n^2+n+1}+2n}=
    =\cos \pi \cdot \frac{1+\frac{1}{n}}{\sqrt{4+\frac{1}{n}+\frac{1}{n^2}}+2}. De aici, limita ceruta este \cos \pi \cdot \frac{1}{4}=\frac{\sqrt 2}{2}.

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