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Andreea09
Pe: 2013-05-26T18:20:26+03:00 In: In:

Hospital

Sa se calculeze limitele de functii:

a) lim (x->2) ( x^2-4)ctg [(pi*x)/2]

b)lim (x->infinit) [x-x^2*ln(1+1/x)]

c)lim (x->infinit) [ln(e^x+x)\ln(e^x-x)];

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1 raspuns

  1. andu_flavius95
    maestru (V)

        \[\begin{array}{l} a)\; {\lim }\limits_{x \to 2} \left( {{x^2} - 4} \right)\cot \frac{{\pi x}}{2} = \limits^{0 \cdot \infty } {\lim }\limits_{x \to 2} \frac{{\left( {{x^2} - 4} \right)}}{{\tan \frac{{\pi x}}{2}}} = \limits^{\frac{0}{0}} {\lim }\limits_{x \to 2} \frac{{2x}}{{\frac{{\frac{\pi }{2}}}{{{{\cos }^2}\frac{{\pi x}}{2}}}}} = \frac{4}{{\frac{\pi }{2}}} = \frac{8}{\pi }.\\ b)\; {\lim }\limits_{x \to \infty } \left( {x - {x^2}\ln \left( {1 + \frac{1}{x}} \right)} \right) = \; {\lim }\limits_{x \to \infty } {x^2}\left( {\frac{1}{x} - \ln \left( {\frac{{x + 1}}{x}} \right)} \right) = {\lim }\limits_{x \to \infty } \frac{{\frac{1}{x} - \ln \left( {\frac{{x + 1}}{x}} \right)}}{{\frac{1}{{{x^2}}}}} = {\lim }\limits_{x \to \infty } \frac{{ - \frac{1}{{{x^2}}} + \frac{1}{{x\left( {x + 1} \right)}}}}{{ - \frac{2}{{{x^3}}}}} = ... = \frac{1}{2}\end{array}\]

    Iar pt c) este mai convenabil sa aplici metoda factorului comun deoarece aplicarea l’Hospital duce la ciclicitate in calcule.

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