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gefest
user (0)
Pe: 2012-01-30T19:42:15+02:00 In: In:

calcul limita de sir

\lim_{n\to\infty}\left\(\sqrt{n^2+2n+3}-n\right\)^{n+\sqrt{n}}

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6 raspunsuri

  1. DD
    profesor

    Expresia data se poate scrie si ;lim(n->infinit){[(n^2+2.n+3)-n^2]/[(radical din (n^2+2.n+3))+n]}^(n+radical din n)=lim(n->infinit){(2n+3)/
    [n.((radical din (1+2/n+3/n^2))+1)]}^n.(1+1/(radical din n))=lim(n-=>infinit)[(2n+3)/2n]^n=lim(n->infinit)[1+3/(2n))^n=e^3/2

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  2. gefest
    user (0)

    \lim\dots=e

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  4. DD
    profesor

    lim(n->infinit)[(1+a/n)^n]=e^a – limita importanta pe care elevii sunt obligati s-o stie.

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  5. gefest
    user (0)

    Nu sunt elev. Mie mi-a iesit e, iar raspunsul din manual este e^{\frac13}.

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  6. DD
    profesor

    Regret , dar problema propusa are raspunsul; e^3/2=e.(radical din e).

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  8. gefest
    user (0)

    \lim_{n\to\infty}\left\(\sqrt{n^2+2n+3}-n\right\)^{n+\sqrt{n}}

    \lim_{n\to\infty}\left\(1+\sqrt{n^2+2n+3}-n-1\right\)^{n+\sqrt{n}}

    \lim_{n\to\infty}\left\(1+\frac{n^2+2n+3-n^2-2n-1)}{\sqrt{n^2+2n+3}+n+1}\right\)^{n+\sqrt{n}}

    \lim_{n\to\infty}\left\(1+\frac{2}{\sqrt{n^2+2n+3}+n+1}\right\)^{n+\sqrt{n}}

    \lim_{n\to\infty}\left\(1+\frac{1}{\frac{\sqrt{n^2+2n+3}+n+1}{2}}\right\)^{\frac{\sqrt{n^2+2n+3}+n+1}{2}\cdot\frac{2(n+\sqrt{n})}{\sqrt{n^2+2n+3}+n+1}}

    Acum calculez limita de la exponent:

    \lim_{n\to\infty}\frac{2(n+\sqrt{n})}{\sqrt{n^2+2n+3}+n+1}=\lim_{n\to\infty}\frac{2n\left\(1+\sqrt{\frac{1}{n}}\right)}{n\left\(\sqrt{1+\frac{2}{n}+\frac{3}{n^2}}+1+\frac{1}{n}\right\)}=\lim_{n\to\infty}\frac{2\left\(1+\sqrt{\frac{1}{n}}\right)}{\sqrt{1+\frac{2}{n}+\frac{3}{n^2}}+1+\frac{1}{n}}=\frac22=1

    Nu gasesc nicio greseala?!

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