Given a Riemannian manifold $(M,g)$, and an exponential map $f_\alpha: B_1 \rightarrow M$ which is given by $$f_\alpha(x):=\exp _{x_0}(\alpha x),$$ where $\exp _p$ is the exponential map at $p$, then define $$h_\alpha:=\alpha^{-2} f_\alpha^* g,$$ which is a metric on $B_1$.
We have $$ \begin{aligned} h_{\alpha, i j} & :=h_\alpha\left(\frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j}\right)=\alpha^{-2} g\left(\frac{\partial f_\alpha}{\partial x^i}, \frac{\partial f_\alpha}{\partial x^j}\right)=: \alpha^{-2} g_{i j} \\ \sqrt{h_\alpha} & :=\sqrt{\operatorname{det}\left(h_{\alpha, i j}\right)}=\alpha^{-2 m} \sqrt{\operatorname{det}\left(g_{i j}\right)}=: \alpha^{-2 m} \sqrt{g}, \end{aligned} $$ I wonder why $h_\alpha \rightarrow|d x|^2$ as $\alpha \rightarrow 0$ in $C^{\ell}\left(B_1\right)$ for every $\ell$, where $|d x|^2$ denotes the Euclidean metric.
My attempt: Consider $$ f_\alpha:=\exp _{x_0}(\alpha x): (B_1, h_\alpha) \rightarrow (M,g_\alpha), $$ and $$ \alpha^{-2}f_\alpha^* g_\alpha=h_\alpha. $$ Following REF1, we have $$ d_{g_\alpha}^2\left(\exp _{x 0}(\alpha v), \exp _{x_0}(\alpha w)\right)=|v-w|^2 \alpha^2-\frac{1}{3} R(v, w, w, v) \alpha^4+O\left(\alpha^5\right)=d_{h_\alpha}^2(v,w), $$ then when $\alpha\rightarrow0$ $$\frac{d_{g_\alpha}^2(\exp _{x 0}(\alpha v), \exp _{x_0}(\alpha w)) }{ \alpha^2} \rightarrow|v-w|^2$$ which is distance function in Euclidean space. Then I read REF2, when $\alpha\rightarrow0$ $$ \alpha^{-2}f_\alpha^* g_\alpha=h_\alpha, $$ which is the Euclidean metric.