Esercizi sui Limiti

  • Materia: Esercizi sui Limiti
  • Visto: 1783
  • Data: 22/06/2007
  • Di: Redazione StudentVille.it

$\\lim_{x \\to +\\infty} \\root{3}{2 + x^3} - \\root{3}{1 + 2x^2 + x^3}$

esercizio svolto o teoria

A cura di: Gianni Sammito

Calcolare

 

$\lim_{x \to +\infty} \root{3}{2 + x^3} - \root{3}{1 + 2x^2 + x^3}$

 


Ricordando il prodotto notevole $(a-b)(a^2 + ab + b^2) = a^3 - b^3$, e moltiplicando numeratore e denominatore per $\root{3}{(2+x^3)^2} + \root{3}{(2+x^3)(1 + 2x^2 + x^3)} + \root{3}{(1 + 2x^2 + x^3)^2}$ si ottiene

 

$\lim_{x \to +\infty} (\root{3}{2 + x^3} - \root{3}{1 + 2x^2 + x^3}) \cdot \frac{\root{3}{(2+x^3)^2} + \root{3}{(2+x^3)(1 + 2x^2 + x^3)} + \root{3}{(1 + 2x^2 + x^3)^2}}{\root{3}{(2+x^3)^2} + \root{3}{(2+x^3)(1 + 2x^2 + x^3)} + \root{3}{(1 + 2x^2 + x^3)^2}} =$

$ = \lim_{x \to +\infty} \frac{2 + x^3 - 1 - 2x^2 - x^3}{\root{3}{[x^3(\frac{2}{x^3} + 1)]^2} + \root{3}{x^3 (\frac{2}{x^3} + 1) x^3 (\frac{1}{x^3} + \frac{2}{x} + 1)} + \root{3}{[x^3 (\frac{1}{x^3} + \frac{2}{x} + 1)]^2}}=$

$ = \lim_{x \to +\infty} \frac{1 - 2x^2}{\root{3}{x^6(\frac{2}{x^3} + 1)^2} + \root{3}{x^6 (\frac{2}{x^3} + 1) (\frac{1}{x^3} + \frac{2}{x} + 1)} + \root{3}{x^6 (\frac{1}{x^3} + \frac{2}{x} + 1)^2}}=$

$ = \lim_{x \to +\infty} \frac{1 - 2x^2}{x^2 \root{3}{(\frac{2}{x^3} + 1)^2} + x^2 \root{3}{(\frac{2}{x^3} + 1) (\frac{1}{x^3} + \frac{2}{x} + 1)} + x^2 \root{3}{(\frac{1}{x^3} + \frac{2}{x} + 1)^2}}$

 

Dividendo numeratore e denominatore per $x^2$ si ottiene

 

 $ = \lim_{x \to +\infty} \frac{\frac{1}{x^2} - 2}{\root{3}{(\frac{2}{x^3} + 1)^2} + \root{3}{(\frac{2}{x^3} + 1) (\frac{1}{x^3} + \frac{2}{x} + 1)} + \root{3}{(\frac{1}{x^3} + \frac{2}{x} + 1)^2}} = \frac{-2}{1 + 1 + 1} = - \frac{2}{3}$

 

FINE