Exercitii cu nedeterminari

Licuricy Lun Ian 30, 2012 9:18 pm

As dori daca se poate sa imi spuneti daca am rezolvat bine exercitiile urmatoare:
$\lim\to \infty (\frac{2x+1}{3x^2+4x+1}*\frac{x}{\sqrt{x^2+1}})=\lim\to \infty \frac{x(2+\frac{1}{x})}{x^2(3+\frac{4}{x}+\frac{1}{x^2})}*\frac{x}{\sqrt{x^2(1+\frac{1}{x^2})}}=\lim\to \infty \frac{x(2+\frac{1}{x})}{x^2(3+\frac{4}{x}+\frac{1}{x^2})}=\lim\to \infty \frac{2}{3x}=0$

$\lim x \to \infty (\frac{4x^2+3}{2x^2+6x+1}*\frac{x+\sqrt{x}}{\sqrt{x^2+4}})=\lim x \to \infty(\frac{x^2(4+\frac{3}{x^2})}{x^2(2+\frac{6}{x}+\frac{1}{x^2})}*\frac{x+\sqrt{x}}{\sqrt{x^2(1+\frac{4}{x^2})}})=\lim x \to \infty (\frac{4+\frac{3}{x^2}}{2+\frac{6}{x}+\frac{1}{x^2}}*\frac{x+\sqrt{x}}{x\sqrt{1+\frac{4}{x^2}}})=2$

**Admin** Mar Ian 31, 2012 1:07 am

Da sunt rezolvate corect, felicitări Very Happy

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Exercitii cu nedeterminari

Exercitii cu nedeterminari

Re: Exercitii cu nedeterminari