Solution: To the contrary, let us assume that $\vert f^\prime (l)\vert > 1$. Since $f$ is continuous, we have \[ l = \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} f\left( x_n \right) = f(l). \] Consider, \begin{align*} \left\vert f^\prime (l) \right\vert & = \left\vert \lim_{x \to l} \frac{f(x) - f(l)}{x - l} \right\vert = \lim_{x \to l} \left\vert \frac{f(x) - f(l)}{x - l} \right\vert > 1. \end{align*}

Take $\varepsilon = \frac{\vert f(l) \vert - 1}{2}$. For $\vert x-l \vert \lt \delta $ we have, \begin{align*} \left\vert \left\vert \frac{f(x) - f(l)}{x - l} \right\vert - \left\vert f^\prime (l) \right\vert \right\vert \lt \varepsilon & \implies \left\vert f^\prime (l) \right\vert - \varepsilon \lt \left\vert \frac{f(x) - f(l)}{x - l} \right\vert \lt \left\vert f^\prime (l) \right\vert + \varepsilon \\ & \implies \left\vert \frac{f(x) - f(l)}{x - l} \right\vert > \frac{\left\vert f^\prime (l) \right\vert + 1}{2} > 1. \end{align*} Using the above inequality, if $\vert x-l \vert \lt \delta $ we have \begin{align*} \vert x - l \vert \lt \vert x - l \vert \frac{\left\vert f^\prime (l) \right\vert + 1}{2} \lt \vert f(x) - f(l) \vert . \end{align*}

Since $x_n \to l$ so for the above $\delta $ we can find $n_0 \in \mathbb{N} $ such that for $n \geq n_0$ \[ \left\vert x_n - l \right\vert \lt \delta . \] Thus, for $n \geq n_0$ \[ \vert x_n - l \vert \lt \left\vert f\left( x_n \right) - f(l) \right\vert = \left\vert x_{n+1} - l \right\vert \] In particular, \begin{align*} \vert x_{n_0} - l \vert \lt \vert x_{n_0 + 1} - l\vert \lt \vert x_n - l \vert. \end{align*} Taking limit, we get \begin{align*} \vert x_{n_0} - l \vert \lt \vert x_{n_0 + 1} - l\vert \leq 0, \end{align*} which is a contradiction. Therefore, we must have $\left\vert f^\prime (l) \right\vert \leq 1$.