Calcolare
[math]\lim_{x \to +\infty} >br>
oot{3}{2 + x^3} - >br>
oot{3}{1 + 2x^2 + x^3}[/math]
Ricordando il prodotto notevole
[math](a-b)(a^2 + ab + b^2) = a^3 - b^3[/math]
, e moltiplicando numeratore e denominatore per [math]>br>
oot{3}{(2+x^3)^2} + >br>
oot{3}{(2+x^3)(1 + 2x^2 + x^3)} + >br>
oot{3}{(1 + 2x^2 + x^3)^2}[/math]
si ottiene
[math]\lim_{x \to +\infty} (>br>
oot{3}{2 + x^3} - >br>
oot{3}{1 + 2x^2 + x^3}) \cdot \frac{>br>
oot{3}{(2+x^3)^2} + >br>
oot{3}{(2+x^3)(1 + 2x^2 + x^3)} + >br>
oot{3}{(1 + 2x^2 + x^3)^2}}{>br>
oot{3}{(2+x^3)^2} + >br>
oot{3}{(2+x^3)(1 + 2x^2 + x^3)} + >br>
oot{3}{(1 + 2x^2 + x^3)^2}} =[/math]
[math] = \lim_{x \to +\infty} \frac{2 + x^3 - 1 - 2x^2 - x^3}{>br>
oot{3}{[x^3(\frac{2}{x^3} + 1)]^2} + >br>
oot{3}{x^3 (\frac{2}{x^3} + 1) x^3 (\frac{1}{x^3} + \frac{2}{x} + 1)} + >br>
oot{3}{[x^3 (\frac{1}{x^3} + \frac{2}{x} + 1)]^2}}=[/math]
[math] = \lim_{x \to +\infty} \frac{1 - 2x^2}{>br>
oot{3}{x^6(\frac{2}{x^3} + 1)^2} + >br>
oot{3}{x^6 (\frac{2}{x^3} + 1) (\frac{1}{x^3} + \frac{2}{x} + 1)} + >br>
oot{3}{x^6 (\frac{1}{x^3} + \frac{2}{x} + 1)^2}}=[/math]
[math] = \lim_{x \to +\infty} \frac{1 - 2x^2}{x^2 >br>
oot{3}{(\frac{2}{x^3} + 1)^2} + x^2 >br>
oot{3}{(\frac{2}{x^3} + 1) (\frac{1}{x^3} + \frac{2}{x} + 1)} + x^2 >br>
oot{3}{(\frac{1}{x^3} + \frac{2}{x} + 1)^2}}[/math]
Dividendo numeratore e denominatore per
[math]x^2[/math]
si ottiene
[math] = \lim_{x \to +\infty} \frac{\frac{1}{x^2} - 2}{>br>
oot{3}{(\frac{2}{x^3} + 1)^2} + >br>
oot{3}{(\frac{2}{x^3} + 1) (\frac{1}{x^3} + \frac{2}{x} + 1)} + >br>
oot{3}{(\frac{1}{x^3} + \frac{2}{x} + 1)^2}} = \frac{-2}{1 + 1 + 1} = - \frac{2}{3}[/math]
FINE
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