The argument exploits the [continuity](/page/Continuity) of $f'$ to find a neighbourhood of $p$ on which $|f'|$ is bounded by a constant $k$ lying strictly between $|\lambda|$ and the critical threshold $1$. The [Mean Value Theorem](/theorems/186) then converts this [derivative](/page/Derivative) bound into a geometric estimate on the distance $|f^n(x_0) - p|$: a contraction in Case 1 and an expansion in Case 2. **Step 1: Asymptotic stability ($|\lambda| < 1$).** Choose $k$ with $|\lambda| < k < 1$. Since $f \in C^1(I, I)$ and $|f'(p)| = |\lambda| < k$, there exists $\epsilon > 0$ such that \begin{align*} |f'(x)| \le k \quad \text{for all } x \in (p - \epsilon, p + \epsilon) \cap I. \end{align*} Let $x_0 \in (p - \epsilon, p + \epsilon) \cap I$. By the Mean Value Theorem, there exists $\xi$ between $x_0$ and $p$ such that \begin{align*} |f(x_0) - p| &= |f(x_0) - f(p)| = |f'(\xi)| \, |x_0 - p| \le k \, |x_0 - p|. \end{align*} Since $k |x_0 - p| < |x_0 - p| < \epsilon$, the image $f(x_0)$ remains in $(p - \epsilon, p + \epsilon) \cap I$. Applying the estimate inductively: \begin{align*} |f^n(x_0) - p| \le k^n \, |x_0 - p| \to 0 \quad \text{as } n \to \infty. \end{align*} **Step 2: Instability ($|\lambda| > 1$).** Choose $k$ with $1 < k < |\lambda|$. Continuity of $f'$ provides $\epsilon > 0$ such that \begin{align*} |f'(x)| \ge k \quad \text{for all } x \in (p - \epsilon, p + \epsilon) \cap I. \end{align*} For any $x_0 \in (p - \epsilon, p + \epsilon) \cap I$ with $x_0 \neq p$, the Mean Value Theorem gives \begin{align*} |f(x_0) - p| = |f'(\xi)| \, |x_0 - p| \ge k \, |x_0 - p| > |x_0 - p|. \end{align*} The distance to $p$ grows by at least the factor $k > 1$ at each iterate as long as the orbit remains in $(p - \epsilon, p + \epsilon)$. Since the displacement increases geometrically and the neighbourhood has finite radius $\epsilon$, the orbit must eventually leave $(p - \epsilon, p + \epsilon)$.