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Coloque aqui todas as dúvidas que tiver sobre limites, regra de Cauchy ou L'Hopital, limites notáveis e afins
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calcule o limite

24 jul 2013, 00:55

\(\lim_{x\rightarrow \infty } \left ( \frac{2x^3-3x+5}{4x^5-2} \right )\)

Re: calcule o limite

24 jul 2013, 15:56

\(\lim_{x \to +\infty} \frac{2x^3-3x+5}{4x^5-2}=\)
\(\lim_{x \to +\infty} \frac{2x^3(1-3/(2x)+5/(2x^3))}{4x^5(1-1/(2x^5))}=\)
\(\lim_{x \to +\infty} \frac{(1-3/(2x)+5/(2x^3))}{2x^2(1-1/(2x^5))}=0\)
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