From my exam

$1 - True$

Since the first derivative of $\phi$ is continuous, $\phi$ is continuous and diferentiable in any compact interval of $\mathbb{R}$ . So, by letting $a<b$ be any two fixed points of $\phi$ , applying Lagrange's Theorem in $[a,b]$ comes that there is one $c \in ]a,b[$ such that:

$\phi ^{'}(c)$ $=\frac{\phi(b)-\phi(a)}{b-a}=\frac{b-a}{b-a}=1$

$2 - True$

For every $n \in \mathbb{N} ^{+}$ let $a_{n}=min$ { $x_{n},x_{n+1}$ } and $b_{n}=max$ { $x_{n},x_{n+1}$ }. By applying Lagrange's Theorem to each on of the intervals $[a_{n},b_{n}]$ , similar to (1), we get that there is a $d_{n} \in ]a_{n},b_{n}[$ such that $\phi ^{'}(d_{n})=1$ . Also, if $x_{n} \rightarrow d$ , then $a_{n} \rightarrow d$ and $b_{n} \rightarrow d$ , which implies by the Squeeze Theorem that $d_{n} \rightarrow d$ . Since $\phi ^{'}$ is continuous at $d$ , by the definition of continuity according to Heine, we get that $\phi ^{'}(d)=lim \phi ^{'} (d_{n})=1$ since it is a constant sequence. Then, $\phi ^{'}(d)=1$ .