Dengan menggunakan definisi turunan \( f^{\prime \prime}(x)=\operatorname{Limit}_{h \rightarrow 0} \frac{1(x+n)-1(x)}{h} \) maka hasil turunar \( \begin{array}{lll}\text { dari } f(x)=3 x^{2}-2 x \text { adalah } f^{\prime}(x)=\ldots & \text { C. } 6 x^{2}-2 x \\ \begin{array}{lll}\text { A. } 3 x-2 & \text { B. } 6 x+2 & \text { E. } 3 x-1\end{array}\end{array} \)

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

Using Fundamental Theorem of Calculus: Part One, find the following: 1. Given \( f(t)=\left(2+t^{4}\right)^{5} \) and \( g(x)=\int_{1}^{x}\left(2+t^{4}\right)^{5} d t \), what is \( g^{\prime}(x)= \) ?

\( \frac { n ! } { ( n + 1 ) ^ { n } } = \)

\( \frac { ( n + 1 ) ! } { ( n + 1 ) ^ { n + 1 } } = \frac { n ! ( n + 1 ) } { ( n + 1 ) ^ { n } ( n + 1 ) } = \frac { n ! } { n ^ { n } ( \frac { n + 1 } { n } ) ^ { n } } = \frac { n ! } { n ^ { n } ( 1 + \frac { 1 } { n } ) ^ { n } } \)

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