[proofplan] The proof is a direct comparison of the two difference quotients. Since $g$ is the restriction of $f$ to the line $t \mapsto a + tv$, its quotient at $0$ is exactly the quotient defining the [directional derivative](/page/Directional%20Derivative) of $f$ at $a$ in the direction $v$. Therefore the two limits exist simultaneously and, when they exist, have the same value. [/proofplan] [step:Identify the two difference quotients] Since $0 \in I$, the definition of $g$ gives \begin{align*} g(0) = f(a + 0v) = f(a). \end{align*} For every $t \in I \setminus \{0\}$, the difference quotient for the ordinary derivative of $g$ at $0$ is \begin{align*} \frac{g(t) - g(0)}{t} = \frac{f(a + tv) - f(a)}{t}. \end{align*} The right-hand side is precisely the difference quotient used in the arbitrary-vector definition of the directional derivative $D_v f(a)$. [guided] We compare the definitions at the level of quotients, before taking any limits. The function \begin{align*} g: I \to \mathbb{R}, \quad t \mapsto f(a + tv) \end{align*} is well-defined because the hypotheses state that $a + tv \in U$ for every $t \in I$, and $f$ has domain $U$. Evaluating at $t = 0$ gives \begin{align*} g(0) = f(a + 0v) = f(a). \end{align*} Thus, for every nonzero $t \in I$, the ordinary one-variable difference quotient of $g$ at $0$ is \begin{align*} \frac{g(t) - g(0)}{t} = \frac{f(a + tv) - f(a)}{t}. \end{align*} This is not merely analogous to the directional derivative quotient; it is the same expression. The variable $t$ measures displacement along the line through $a$ in direction $v$, so substituting the line restriction converts the one-variable quotient into the directional quotient exactly. [/guided] [/step] [step:Take the common limit] By definition, \begin{align*} g'(0) = \lim_{t \to 0} \frac{g(t) - g(0)}{t} \end{align*} when this limit exists. By the definition of the directional derivative in direction $v$, \begin{align*} D_v f(a) = \lim_{t \to 0} \frac{f(a + tv) - f(a)}{t} \end{align*} when this limit exists, with $t$ restricted to values for which $a + tv \in U$. Since $I$ is an open interval containing $0$, it supplies such values of $t$ in a neighbourhood of $0$, and the two quotients are equal for all nonzero $t \in I$. Hence the two limits exist if and only if each other exists, and in that case they are equal: \begin{align*} D_v f(a) = g'(0). \end{align*} This proves the theorem. [/step]