I am reading the John Lee's Introduction to Riemannian manifold, p.266~ 271, proof of Theorem 9.1 ( Rotation Index theorem ) and trying to understand some statement.
Let $\gamma : [a,b] \to \mathbb{R}^2$ be a positively oriented curved polygon ( by definition, refer to John lee's book p.265 ). In fifth paragraph of the proof, he argues as follows :
Let $ a \leq t_1 < t_2 \leq b$. Then the fundamental theorem of calculus gives
$$\gamma(t_2)-\gamma(t_1) = \int_{0}^{1}\frac{d}{ds}\gamma(t_1 + s(t_2 - t_1))ds=\int_{0}^{1}\gamma'(t_1+s(t_2-t_1))(t_2-t_1)ds.$$
And thus
$$ \left|\frac{\gamma(t_2)-\gamma(t_1)}{t_2-t_1} -\gamma'(t) \right|\le \int_{0}^{1}|\gamma'(t_1+s(t_2-t_1))-\gamma'(t) | ds.$$
Because $\gamma'$ is uniformly continuous on the compact set $[a,b]$, this last expression can be made as small as desired by taking $(t_1, t_2)$ close to $(t,t)$. It follows that
$$\lim_{(t_1,t_2) \to (t,t), t_1 < t_2} \frac{\gamma(t_2)-\gamma(t_1)}{t_2-t_1}=\gamma'(t). $$
And why this is true? How can we use the uniform continuity of $\gamma'$? Can anyone helps?
According to uniformly continuous, we know that $\forall\epsilon>0$ there exist a $\delta>0$ such that $\forall \tau_{1},\tau_{2}$, if $|\tau_{1}-\tau_{2}|\leq\delta$, then we have $|\gamma'(\tau_{1})-\gamma'(\tau_{2})|\leq\epsilon$.
In your question, when $t_{1},t_{2}$ close enough, that is to say, $|t_{1}+s(t_{2}-t_{1})-t|\leq\delta$, so we have$|\gamma'(t_{1}+s(t_{2}-t_{1}))-\gamma'(t)|\leq\epsilon$, you integral domain about $t$ is $[0,1]$ finite, so we finish the argument!