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Pe: 2017-07-31T16:38:16+03:00 In: In:

limita

problema 3 va roog

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5 raspunsuri

  1. profesor

    \lim_{x\to 2}\frac{\sqrt[3]{12x+3}-\sqrt{x+7}}{x-2}=\lim_{x\to 2}\left ( \frac{\sqrt[3]{12x+3}-3}{x-2}-\frac{\sqrt{x+7}-3}{x-2} \right )=\\=f'(2)-g'(2),\;unde\;f'(x)=\frac{12}{3\sqrt[3]{(12x+3)^2}},\;g'(x)=\frac{1}{2\sqrt{x+7}}

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  2. profesor

    Oricare ar fi a, b pozitive, are loc identitatea arctga-arctgb=arctg\frac{a-b}{1+ab}.
    Limita devine \lim_{x\to \infty }x\cdot arctg\,u(x)=\lim_{x\to \infty }\frac{arctg\,u(x)}{u(x)}\cdot (xu(x)).
    Primul factor are limita 1, pentru că u(x) are limita 0, iar al doilea – 1/2.

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  4. user (0)

    ghioknt post_id=109458 time=1501527918 user_id=18683 wrote:
    \lim_{x\to 2}\frac{\sqrt[3]{12x+3}-\sqrt{x+7}}{x-2}=\lim_{x\to 2}\left ( \frac{\sqrt[3]{12x+3}-3}{x-2}-\frac{\sqrt{x+7}-3}{x-2} \right )=\\=f'(2)-g'(2),\;unde\;f'(x)=\frac{12}{3\sqrt[3]{(12x+3)^2}},\;g'(x)=\frac{1}{2\sqrt{x+7}}


    multumesc! si fara derivate nu s-ar putea calcula limita?

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  5. profesor

    Ba da, amplificând fiecare raport cu expresiile conjugate ale numărătorilor.

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  6. profesor

    4b))lim┬(x→∞)⁡((e^x+x)^(n+1)-e^(n+1)x )/(x.e^nx )=〖lim⁡[〗┬(x→∞) (e^(n+1)x+C_( n+1)^1.x.e^nx+C_(n+1)^2.x^2.e^(n-1)x+⋯+C_(n+1)^(n+1).x^((n+1) ) )-
    -e^(n+1)x]/x.e^nx=lim┬(x→∞) ((n+1)+C_(n+1)^2 (x/eˆx )+ +C_(n+1)^(n+1) (x^n/e^nx ))=n+1
    OBS;lim(x->Inf)(x/e^x)^k=0,pentru orjce k ce apartin lui N

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