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