Let $\alpha \in \mathbb{R}$，$\ 0<\alpha <2 \ $, the fractional laplacian is defined as:
\begin{equation}\label{def} (-\Delta)^{\frac{\alpha}{2}} u (x) := C_{n,\alpha} \lim\limits_{\epsilon \rightarrow 0^+} \int_{\mathbb{R}^n \verb|\| B_{\epsilon}(x)} \frac{u(x)-u(z)}{ |x-z|^{n+ \alpha}} dz , \end{equation}
When $u \in C_{loc} ^{1,1}(\mathbb{R}^n) \cap L_{\alpha} (\mathbb{R}^n)= \left\{ u \in L_{loc} ^1(\mathbb{R}^n)| \int_{\mathbb{R}^n} \frac{|u(x)|}{1+|x|^{n+\alpha}} dx < \infty \right\}$, the integral is well defined.
We can seperate the integral into two parts.
By Taylor expansion and $ u \in C_{loc} ^{1,1}(\mathbb{R}^n)$,which means $\nabla u$ is Lipschitz continuous in every compact subset in $\Omega$ , we have $$u(x)-u(z)=-\nabla u(x)(z-x) + O(|x-z|^2),$$
So
$$ \lim\limits_{\epsilon \rightarrow 0^+} \int_{B_1(x) \verb|\| B_{\epsilon}(x)} \frac{u(x)-u(z)}{ |x-z|^{n+ \alpha}} dz $$ $$= \lim\limits_{\epsilon \rightarrow 0^+} \int_{B_1(x) \verb|\| B_{\epsilon}(x)} \frac{-\nabla u(x)(z-x) + O(|x-z|^2)}{|x-z|^{n+ \alpha}} dz $$
By the symmetry of the integral
$$= \lim\limits_{\epsilon \rightarrow 0^+} \int_{B_1(x) \verb|\| B_{\epsilon}(x)} \frac{O(|x-z|^2)}{|x-z|^{n+ \alpha}} dz $$ $$\leq C \lim\limits_{\epsilon \rightarrow 0^+} \int_{B_1(x) \verb|\| B_{\epsilon}(x)} \frac{1}{|x-z|^{n+ \alpha -2}} dz < \infty.$$
I'm confused about the Cauchy principal here. Why do we need Cauchy princial? I think the integral converges in usual sense.
You need the Cauchy principle value in order to show that $$\int_{B_1(x)} \frac{\nabla u (x) \cdot (y-x)}{\vert x-y\vert^{n+\alpha}} \, dy=0. $$ The integral on the right hand side is not integrable in the usual sense (depending on the value of $\alpha$). To see this let's just look at the case $n=1$. Then the integral is $$u' (x)\int_{x-1}^{x+1} \frac{ (y-x)}{\vert x-y\vert^{1+\alpha}} \, dy =u' (x)\int_{-1}^1 \frac{z}{\vert z \vert^{1+\alpha}}\, dz.$$ If we do not use the Cauchy principle value then by definition $$ \int_{-1}^1 \frac{z}{\vert z \vert^{1+\alpha}}\, dz = \lim_{a\to 0^+}\int_{-1}^{-a} \frac{z}{\vert z \vert^{1+\alpha}}\, dz + \lim_{b\to0^+}\int_b^1 \frac{z}{\vert z \vert^{1+\alpha}}\, dz$$ provided both these limits exist. For $0<\alpha<1$ these limits do exist so we actually don't need to use the Cauchy principle value in this case; however, if $\alpha\geqslant 1$ then neither limit exists, so the integral is undefined. Now if we use the Cauchy principle value $$\mathrm{P.V.}\,\int_{-1}^1 \frac{z}{\vert z \vert^{1+\alpha}}\, dz =\lim_{\varepsilon\to0^+}\bigg [ \int_{-1}^{-\varepsilon} \frac{z}{\vert z \vert^{1+\alpha}}\, dz+\int_{\varepsilon}^1 \frac{z}{\vert z \vert^{1+\alpha}}\, dz \bigg ]=0 $$ by symmetry.
If we return to your answer, you need to take the Cauchy principle value since in general $$\int_{B_1(x)} \frac{\nabla u (x) \cdot (y-x)}{\vert x-y\vert^{n+\alpha}} \, dy \neq \lim_{\varepsilon \to 0^+ }\int_{B_1(x)\setminus B_\varepsilon(x)} \frac{\nabla u (x) \cdot (y-x)}{\vert x-y\vert^{n+\alpha}} \, dy.$$
As a side note, I noticed you wrote $$\tag{$\ast$}\lim_{\varepsilon \to 0^+}\int_{B_1(x)\setminus B_\varepsilon(x)} \frac{dy}{\vert x-y\vert^{n+\alpha}} <+\infty $$ which is not true. You need to take your Taylor series to include the quadratic term so that at the end, instead of $(\ast)$, you get $$\lim_{\varepsilon \to 0^+}\int_{B_1(x)\setminus B_\varepsilon(x)} \frac{\vert x - y \vert^2}{\vert x-y\vert^{n+\alpha}} \, dy $$ which is finite.