A connection is the limit of the newton quotient of the parallel transport Let $E\rightarrow M$ be a vector bundle with connection $\nabla$. Denote by $\Pi_{\gamma(t_{0})}^{\gamma(t_{1})}:E_{\gamma(t_{0})}\rightarrow E_{\gamma(t_{1})}$ the parallel transport map along the curve $\gamma$. 
Show that for any $X\in \Gamma(TM)$ and $s\in\Gamma(E) $
$$(\nabla_{X}s)(p)=\lim_{\delta\rightarrow 0}\dfrac{\Pi_{\gamma(\delta)}^{\gamma(0)}(s(\gamma(\delta)))-s(\gamma(0))}{\delta}$$
where $\gamma$ is any curve in $M$ such that $\gamma(0)=p$ and $\gamma'(0)=X(p)$.
So I have no idea how to do this should any tips for this?
 A: This shows you, that the connection is completely encoded in it's parallel transport and in this encoding, it's nothing than the regular derivative.
Let $\xi_1,...,\xi_n$ be a basis of $E_p$ and $s_1,...,s_n$ be the unique parallel sections along $\gamma$ with $s_i(0)=\xi_p$. Define $\tilde{S}=S\circ\gamma$ and let $\tilde{S}(t)=\sum\sigma^j(t)s_j(t)$. The following calculation then does the job:
\begin{align*}
\nabla_{X_p}S = \tilde S'(0) &= \nabla_t\big(\sum_j\sigma^js_j\big)(0)\\
&= \sum_j\big((\sigma^j)'s_j + \underbrace{\sigma^j\nabla_ts_j}_{\equiv\ 0}\big)(0)\\
&= \sum_j\bigg(\lim_{t\to0}\frac{\sigma^j(t)-\sigma^j(0)}{t}\bigg)s_j(0)\\
&= \lim_{t\to 0}\frac{1}{t}\Big(\sum_j\sigma^j(t)s_j(0) - \sigma^j(0)s_j(0)\Big)\\
&= \lim_{t\to 0}\frac{1}{t}\Big(\sum_j\sigma^j(t)P_{t,0}^\gamma(s_j(t))-\sigma^j(0)s_j(0)\Big)\\
&= \frac{1}{t}\Big(P_{t,0}^\gamma\big(\sum_j\sigma^j(t)s_j(t)\big)-\sum_j\sigma^j(0)s_j(0)\Big)\\
&= \lim_{t\to0}\frac{P_{t,0}^\gamma(\tilde S(t))-\tilde S(0)}{t}
\end{align*}
I think you can handle the few changes of notation. Otherwise feel free to ask.
