Show that if $\gamma_{n} \colon [0, 1] \rightarrow M$ is a sequence of geodesics in a Riemannian manifold $M$ such that $\gamma(t) = \lim_{n \rightarrow \infty} \gamma_{n}$, then $\gamma$ is a geodesic and for each $\epsilon > 0$ there is $N \in \mathbb{N}$ such that for all $n > N$ we have $d_{g}(\gamma_{n}(t), \gamma(t)) < \epsilon$.
To show that $\gamma(t) = \lim_{n \rightarrow \infty} \gamma_{n}$ is a geodesic, we must show that $\frac{\nabla \dot{\gamma}(t)}{dt} = 0$. But
\begin{equation*} \begin{aligned} \frac{\nabla \dot{\gamma}(t)}{dt} &= \frac{\nabla \dot{\lim_{n \rightarrow \infty} \gamma_{n}}(t)}{dt} \\ &= \frac{\nabla \lim_{n \rightarrow \infty} \dot{\gamma_{n}}(t)}{dt} \\ &= \lim_{n \rightarrow \infty} \frac{\nabla \dot{\gamma_{n}}(t)}{dt} \\ &= 0. \end{aligned} \end{equation*}
The issue I have is that I don't know how to justify the derivative of the limit is the limit of the derivatives (i. e. $\frac{d}{dt}\lim_{n \rightarrow \infty} \gamma_{n} = \lim_{n \rightarrow \infty} \frac{d}{dt}\gamma_{n}$) and interchanging the limit with the connection $\nabla$ (i. e. $\frac{\nabla \lim_{n \rightarrow \infty} \dot{\gamma}(n)}{dt} = \lim_{n \rightarrow \infty} \frac{\nabla \dot{\gamma}_{n}}{dt}$).
I am not sure how to proceed to show that pointwise convergence of geodesics implies uniform convergence.
Suppose that image of geodesic lies in a coordinate neighborhood. Then by compactness of the domain, convergence is uniform. In general case cover image by finitely many coordinate neighborhoods and proof follows.