a. \( \begin{array}{l} f(x)=x^{3}+x+2(0.2) \\ \text { b. } f(x)=\sqrt{20-x^{2}}(2.4) \\ \text { c } f(x)=\left\{\begin{array}{l}x+1+x>3 \\ x^{2}-6 \text { if } x \leqslant 3\end{array}\right. \\ \text { d. } f(x)=\left(-x^{3}\right) \sqrt{x+2)} \cdot(-1.2)\end{array} \)

\( \sum_{n=1}^{\infty} \frac{6}{4^{n}-7} \) (use Direct Companison)

\( \begin{array}{ll}\text { (2) } \int \frac{d x}{7-3 x} & \text { (3) } \int \sqrt[3]{x+5} d x \\ \text { (5) } \int \frac{d x}{(2 x-5)^{2}} & \text { (6) } \int \sqrt[3]{(5-3 x)^{2}} d x\end{array} \)

\( \sum_{n=0}^{\infty} \frac{\cos (n \pi)}{n+5} \) converge conditionty, absolutedy, or diverge?

\[ \frac{n!}{n^{n}} \leq \frac{1}{2^{n-1}}, \quad \forall n \in \mathbb{N}, \quad \text { dopo aver dimostrato per induzione la validità della disuguaglianza } \] verificare che la successione \( a_{n}=\frac{n!}{n^{n}} \) converge a zero.

\( \frac { ( n + 1 ) ! } { ( n + 1 ) ^ { n + 1 } } \leq \frac { 1 } { 2 ^ { n } } \)