Let $f : \mathbb{R} \to \mathbb{R}$ be a D4364: Real function such that

(i)	\begin{equation} \forall \, x \in \mathbb{R} : f(x) = x^2 \end{equation}

Let $L : \mathbb{R} \to \mathbb{R}$ be a D4364: Real function such that

(i)	$x_0 \in \mathbb{R}$ is a D993: Real number
(ii)	\begin{equation} \forall \, x \in \mathbb{R} : L(x) = 2 x_0 x \end{equation}

Fix $x, x_0 \in \mathbb{R}$ such that $x \neq x_0$. Then \begin{equation} \begin{split} \frac{|f(x) - f(x_0) - L(x - x_0)|}{|x - x_0|} & = \frac{|x^2 - x^2_0 - 2 x_0 (x - x_0)|}{|x - x_0|} \\ & = \frac{|(x + x_0) (x - x_0) - 2 x_0 (x - x_0)|}{|x - x_0|} \\ & = \frac{| x - x_0 ||(x + x_0) - 2 x_0 |}{|x - x_0|} \\ & = |(x + x_0) - 2 x_0 | \\ & \to |2 x_0 - 2 x_0| \\ & = 0 \end{split} \end{equation} as $x \to x_0$. The claim follows. $\square$