Ciao, considerando che il parametro alfa, se non ci sono condizioni date dal testo, si presuppone un numero reale qualsiasi, dovendo essere la base di un esponenziale sempre positiva, allora il limite dovrà essere solo per x->0+ poichè dovrà essere:
\[x - sen(x) > 0\;\;\; \to \;\;\;x > sen(x)\;\;\; \to \;\;\;x > 0\]
dunque il limite sinistro per x -> 0- non esiste!
Considerando il limite solo limite destro e sviluppando con Taylor al 2° termine il seno, il coseno e la radice (senza l'uso degli o-piccoli essendo in questo caso il limite molto semplice), si ha:
\[\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\cos (\alpha x) - \sqrt {1 + {x^2}} - {\alpha ^2}{x^2}}}{{{{(x - sen(x))}^\alpha }}} = \]
\[ = \left\{ \begin{array}{l}
\alpha = 0\;\;\; \to \;\;\;\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\cos (0) - \sqrt {1 + {x^2}} - 0}}{{{{(x - sen(x))}^0}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{1 - \sqrt {1 + {x^2}} }}{1} = 0\\
\alpha < 0\;\;\; \to \;\;\;\mathop {\lim }\limits_{x \to {0^ + }} (\cos (\alpha x) - \sqrt {1 + {x^2}} - {\alpha ^2}{x^2}){(x - sen(x))^{ - \alpha }} = 0\\
\alpha > 0\;\;\; \to \;\;\;\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\cos (\alpha x) - \sqrt {1 + {x^2}} - {\alpha ^2}{x^2}}}{{{{(x - sen(x))}^\alpha }}} = [\frac{0}{0}]
\end{array} \right.\]
\[\alpha > 0\;\;\; \to \;\;\;\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\cos (\alpha x) - \sqrt {1 + {x^2}} - {\alpha ^2}{x^2}}}{{{{(x - sen(x))}^\alpha }}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{[1 - \frac{1}{2}{\alpha ^2}{x^2}] - [1 + \frac{1}{2}{x^2}] - {\alpha ^2}{x^2}}}{{{{(x - [x - \frac{1}{6}{x^3}])}^\alpha }}} = \]
\[ = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{1 - \frac{1}{2}{\alpha ^2}{x^2} - 1 - \frac{1}{2}{x^2} - {\alpha ^2}{x^2}}}{{{{(x - x + \frac{1}{6}{x^3})}^\alpha }}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{( - \frac{1}{2} - 1){\alpha ^2}{x^2} - \frac{1}{2}{x^2}}}{{{{(\frac{1}{6}{x^3})}^\alpha }}} = \]
\[ = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{ - \frac{3}{2}{\alpha ^2}{x^2} - \frac{1}{2}{x^2}}}{{{{(\frac{1}{6})}^\alpha }{x^{3\alpha }}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{( - \frac{3}{2}{\alpha ^2} - \frac{1}{2}){x^2}}}{{{{(\frac{1}{6})}^\alpha }{x^{3\alpha }}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{( - \frac{3}{2}{\alpha ^2} - \frac{1}{2})}}{{{{(\frac{1}{6})}^\alpha }{x^{3\alpha - 2}}}} = \]
\[ = \left\{ {\begin{array}{*{20}{l}}
{3\alpha - 2 > 0\;\;\; \to \;\;\;\alpha > \frac{2}{3}\;\;\; \to \;\;\;\mathop {\lim }\limits_{x \to {0^ + }} \frac{{( - \frac{3}{2}{\alpha ^2} - \frac{1}{2})}}{{{{(\frac{1}{6})}^\alpha }{x^{3\alpha - 2}}}} = \frac{{( - \frac{3}{2}{\alpha ^2} - \frac{1}{2})}}{{{0^ + }}} = - \infty }\\
{3\alpha - 2 < 0\;\;\; \to \;\;\;0 < \alpha < \frac{2}{3}\;\;\; \to \;\;\;\mathop {\lim }\limits_{x \to {0^ + }} \frac{{( - \frac{3}{2}{\alpha ^2} - \frac{1}{2})}}{{{{(\frac{1}{6})}^\alpha }{x^{3\alpha - 2}}}} = \mathop {\lim }\limits_{x \to {0^ + }} 6( - \frac{3}{2}{\alpha ^2} - \frac{1}{2}){x^{ - (3\alpha - 2)}} = 0}\\
{3\alpha - 2 = 0\;\;\; \to \;\;\;\alpha = \frac{2}{3}\;\;\; \to \;\;\;\mathop {\lim }\limits_{x \to {0^ + }} \frac{{( - \frac{3}{2}{\alpha ^2} - \frac{1}{2})}}{{{{(\frac{1}{6})}^\alpha }{x^{3\alpha - 2}}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{( - \frac{3}{2}{{(\frac{2}{3})}^2} - \frac{1}{2})}}{{{{(\frac{1}{6})}^{\frac{2}{3}}}{x^0}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{ - 7\frac{1}{6}}}{{{{(\frac{1}{6})}^{\frac{2}{3}}}}} = - 7{{(\frac{1}{6})}^{^{\frac{1}{3}}}} = - 7\sqrt[3]{{\frac{1}{6}}}}\\
{\alpha \le 0\;\;(da\;quanto\;visto\;inizialmente)\;\; \to \;\;\;\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\cos (\alpha x) - \sqrt {1 + {x^2}} - {\alpha ^2}{x^2}}}{{{{(x - sen(x))}^\alpha }}} = 0}
\end{array}} \right.\]
Saluti.
