N. 4 LIMITI CON TAYLOR

Determinare gli eventuali α∈ℝ per i quali il limite indicato esiste finito e in tal caso calcolarlo:

$lim_{x \to 0} \frac{\frac{1}{2} \cdot \ln (1 + x^{2}) - {(\sin x)}^{2} + \frac{1}{2} \cdot \tan (x^{2})}{x^{α}}$

Sviluppi di Taylor:

$\ln (1 + x^{2}) = x^{2} - \frac{x^{4}}{2} + o (x^{4})$

$\sin x = x - \frac{x^{3}}{6} + o (x^{4})$

$\tan (x^{2}) = x^{2} + \frac{x^{6}}{3} + o (x^{8})$

Sviluppo del numeratore:

$N (x) = \frac{1}{2} \cdot \ln (1 + x^{2}) - {(\sin x)}^{2} + \frac{1}{2} \cdot \tan (x^{2}) = \frac{1}{2} \cdot [x^{2} - \frac{x^{4}}{2} + o (x^{4})] - {[x - \frac{x^{3}}{6} + o (x^{4})]}^{2} + \frac{1}{2} \cdot [x^{2} + \frac{x^{6}}{3} + o (x^{8})] =$

$= \frac{x^{2}}{2} - \frac{x^{4}}{4} + o (x^{4}) - x^{2} - \frac{x^{6}}{36} + o (x^{8}) + \frac{x^{4}}{3} + o (x^{5}) + o (x^{7}) + \frac{x^{2}}{2} + \frac{x^{6}}{6} + o (x^{8}) = \frac{1}{12} x^{4} + \frac{5}{36} x^{6} + o (x^{4})$

In conclusione:

$lim_{x \to 0} \frac{\frac{1}{2} \cdot \ln (1 + x^{2}) - {(\sin x)}^{2} + \frac{1}{2} \cdot \tan (x^{2})}{x^{α}} = lim_{x \to 0} \frac{\frac{1}{12} x^{4} + \frac{5}{36} x^{6} + o (x^{4})}{x^{α}} = lim_{x \to 0} \frac{\frac{1}{12} + \frac{5}{36} x^{2} + o (1)}{x^{α - 4}}$

Possono presentarsi tre casi:

${\begin{matrix} α < 4 \to α - 4 < 0 \to lim_{x \to 0} \frac{\frac{1}{12} + \frac{{5 x}^{2}}{12} + o (1)}{x^{α - 4}} = 0 \\ α = 4 \to α - 4 = 0 \to lim_{x \to 0} \frac{\frac{1}{12} + \frac{{5 x}^{2}}{12} + o (1)}{x^{0}} = \frac{1}{12} \\ α > 4 \to α - 4 > 0 \to lim_{x \to 0} \frac{\frac{1}{12} + \frac{5 x^{2}}{12} + o (1)}{x^{α - 4}} = + \infty (non interessa) \end{matrix}$