I'm going through some differential geometry exercises (from Kristopher Tapp's Differential Geometry of Curves and Surfaces) I worked on a while ago, and realised I missed a detail in Part (2) of the following:
Let $\boldsymbol\gamma: I \to \mathbb R^n$ be a unit-speed curve. Let $t_0 \in I$ and $\kappa(t_0) \neq 0$. For sufficiently small $h > 0$, prove that the three points $\boldsymbol\gamma(t_0 - h), \boldsymbol\gamma(t_0)$, and $\boldsymbol\gamma(t_0 + h)$ are not collinear, so there is a unique plane $P_h$ containing them and a unique circle $C_h$ containing them. Precisely formulate and prove the following:
(1) As $h \to 0$, $P_h$ converges to the osculating plane (translated to $\boldsymbol\gamma(t_0)$).
(2) As $h \to 0$, $C_h$ converges to the osculating circle (translated to $\boldsymbol\epsilon(t_0)$).
HINT: For (1), use the Taylor approximation formulas from this section. For (2), for fixed $h$, let $\mathbf p(h)$ denote the center of $C_h$, and define $$f(s) = \lvert\boldsymbol\gamma(t_0 + s) - \mathbf p(h)\rvert^2.$$ Since $f(-h) = f(0) = f(h)$, the mean value theorem says that there exist $\delta_1 \in (-h, 0)$ and $\delta_2 \in (0, h)$ with $f'(\delta_1) = f'(\delta_2) = 0$, and then that there exists $\epsilon \in (\delta_1, \delta_2)$ with $f''(\epsilon) = 0$, which becomes \begin{align} 0 = f'(\delta_i) &= 2 \langle\boldsymbol\gamma'(t_0 + \delta_i), \boldsymbol\gamma(t_0 + \delta_i) - \mathbf p(h)\rangle \textit{ (for $i \in \{1, 2\}$),}\\ 0 = f''(\epsilon) &= 2 \langle\boldsymbol\gamma''(t_0 + \epsilon), \boldsymbol\gamma(t_0 + \epsilon) - \mathbf p(h)\rangle + 2 \lvert\boldsymbol\gamma'(t_0 + \epsilon)\rvert^2. \end{align} Now consider the limit as $h \to 0$.
In the exercise, $\kappa$ is the curvature function and $\boldsymbol\epsilon$ the center of the osculating circle. Further, the text treats all curves as smooth.
I interpreted the statement as saying that (in addition to the statement in Part (1)) $\lim_{h \to 0^+}\mathbf p(h) = \boldsymbol\epsilon(t_0)$. What I actually showed, however, was that if $\lim_{h \to 0^+}\mathbf p(h) = \mathbf L$, then $\mathbf L = \boldsymbol\epsilon(t_0)$:
We define $$f(s) = \lvert\boldsymbol\gamma(t_0 + s) - \mathbf p(h)\rvert^2.$$ Since $f$ is smooth and $f(-h) = f(0) = f(h)$, the mean value theorem says that there exist $\delta_1 \in (-h, 0)$ and $\delta_2 \in (0, h)$ with $f'(\delta_1) = f'(\delta_2) = 0$, and then that there exists $\epsilon \in (\delta_1, \delta_2)$ with $f''(\epsilon) = 0$, which becomes \begin{align} \langle\boldsymbol\gamma'(t_0 + \delta_i), \boldsymbol\gamma(t_0 + \delta_i) - \mathbf p(h)\rangle &= 0\\ \langle\boldsymbol\gamma''(t_0 + \epsilon), \boldsymbol\gamma(t_0 + \epsilon) - \mathbf p(h)\rangle &= -1. \end{align} Now, as $h \to 0$, $\delta_i, \epsilon \to 0$, and so if $\lim_{h \to 0^+}\mathbf p(h) = \mathbf L$, \begin{align} \langle\boldsymbol\gamma'(t_0), \boldsymbol\gamma(t_0) - \mathbf L\rangle &= 0\\ \langle\boldsymbol\gamma''(t_0), \boldsymbol\gamma(t_0) - \mathbf L\rangle &= -1. \end{align} This tells us that $\mathbf L - \boldsymbol\gamma(t_0) \perp \boldsymbol\gamma'(t_0)$. Further, since $\mathbf p(h)$ lies in $P_h$, $\mathbf L$ lies in the translated osculating plane, and so $\mathbf L - \boldsymbol\gamma(t_0)$ lies in the osculating plane. Thus, $\mathbf L - \boldsymbol\gamma(t_0) \parallel \boldsymbol\gamma''(t_0)$. We can then conclude that $\mathbf L - \boldsymbol\gamma(t_0)$ points in the direction of $\boldsymbol\gamma''(t_0)$ and that $\lvert\mathbf L - \boldsymbol\gamma(t_0)\rvert = \frac{1} {\kappa(t_0)}$, i.e. $\mathbf L - \boldsymbol\gamma(t_0) = \frac{1} {\kappa(t_0)} \mathfrak n$ (where $\mathfrak n$ is the unit normal vector). Thus $\mathbf L = \boldsymbol\epsilon(t_0)$.
However, this doesn't rule out the possibility that $\mathbf p(h)$ doesn't have a limit as $h \to 0$. I thought to try and finish the proof I should show that $\mathbf p$ is smooth with bounded derivative, and hence uniformly continuous. In doing so, however, I found $\mathbf p(h) - \boldsymbol\gamma(t_0)$ in terms of $\mathbf u_1(h) := \boldsymbol\gamma(t_0 + h) - \boldsymbol\gamma(t_0)$ and $\mathbf u_2(h) := \boldsymbol\gamma(t_0 - h) - \boldsymbol\gamma(t_0)$, and a direct calculation of the limit seems to suggest that the limit is $\boldsymbol\gamma(t_0)$ instead.
To find $\mathbf p(h) - \boldsymbol\gamma(t_0)$, I set $\mathbf p(h) - \boldsymbol\gamma(t_0) = \alpha(h) \mathbf u_1(h) + \beta(h) \mathbf u_2(h)$. I then used the condition that $\langle\mathbf p(h) - \boldsymbol\gamma(t_0 + h), \mathbf p(h) - \boldsymbol\gamma(t_0 + h)\rangle = \langle\mathbf p(h) - \boldsymbol\gamma(t_0), \mathbf p(h) - \boldsymbol\gamma(t_0)\rangle = \langle\mathbf p(h) - \boldsymbol\gamma(t_0 - h), \mathbf p(h) - \boldsymbol\gamma(t_0 - h)\rangle$ along with the facts that $\mathbf p(h) - \boldsymbol\gamma(t_0 + h) = (\alpha(h) - 1) \mathbf u_1(h) + \beta(h) \mathbf u_2(h)$ and $\mathbf p(h) - \boldsymbol\gamma(t_0 - h) = \alpha(h) \mathbf u_1(h) + (\beta(h) - 1) \mathbf u_2(h)$ to solve for $\alpha(h), \beta(h)$. As a result, I found $$\mathbf p(h) - \boldsymbol\gamma(t_0) = \frac{\lvert\mathbf u_2(h)\rvert^2 \langle\mathbf u_1(h), \mathbf u_1(h) - \mathbf u_2(h)\rangle \mathbf u_1(h) + \lvert\mathbf u_1(h)\rvert^2 \langle\mathbf u_2(h) - \mathbf u_1(h), \mathbf u_2(h)\rangle \mathbf u_2(h)}{2 (\lvert\mathbf u_1(h)\rvert^2 \lvert\mathbf u_2(h)\rvert^2 - \langle\mathbf u_1(h), \mathbf u_2(h)\rangle^2)}.$$ But, since $\lim_{h \to 0^+}\mathbf u_1(h) / h = \boldsymbol\gamma'(t_0)$ and $\lim_{h \to 0^+}\mathbf u_2(h) / h = -\boldsymbol\gamma'(t_0)$, this should imply that $$\lim_{h \to 0^+}(\mathbf p(h) - \boldsymbol\gamma(t_0)) = \lim_{h \to 0^+}\frac{\left\lvert\frac{\mathbf u_2(h)}{h}\right\rvert^2 \left\langle\frac{\mathbf u_1(h)}{h}, \mathbf u_1(h) - \mathbf u_2(h)\right\rangle \frac{\mathbf u_1(h)}{h} + \left\lvert\frac{\mathbf u_1(h)}{h}\right\rvert^2 \left\langle\mathbf u_2(h) - \mathbf u_1(h), \frac{\mathbf u_2(h)}{h}\right\rangle \frac{\mathbf u_2(h)}{h}}{2 \left(\left\lvert\frac{\mathbf u_1(h)}{h}\right\rvert^2 \left\lvert\frac{\mathbf u_2(h)}{h}\right\rvert^2 - \left\langle\frac{\mathbf u_1(h)}{h}, \frac{\mathbf u_2(h)}{h}\right\rangle^2\right)} = \mathbf0.$$ Am I going wrong somewhere?
