FIRST proof. WLOG let $a=1$. Let $-u(z)$ be your Poisson integral, (in the numerator of your formula) it is a negative harmonic function in the disc, continuous in the closed disc, $u(1)=0$, and negative at every other point of the circle. Let $M(r)=max_{|z|=r}u(z),\; 0\leq r\leq 1$. This is a strictly negative, increasing function on $(0,1)$, and $M(1)=1$. It is known that $M(r)$ is convex with respect to the logarithm, that is $$r\frac{dM(r)}{dr}$$ is increasing. This is called Hadamard's Three Circles Theorem. Thus $M'(1)>0$. Now as $u(re^{i\theta})$ is an even function of $\theta$ decreasing on $(0,\pi)$, we conclude that $M(r)=u(r)$. This means that there exists a sequence $z_k\to 1$,such that $z_k<1$, and
$$\frac{u(1)-u(z_k)}{1-z_k}\; \quad\to c>0.$$ So on this sequence your limit $$\frac{-u(z_k)}{|1-z_k|}\quad =c.$$ On the other hand it is clear that your limit is zero on sequences which tend to $1$ "tangentially$, that is very close to the circle. So the limit does not exist. SECOND proof. WLOG$a=1$. Let$v$be your Poisson integral in the numerator. Let$w$be the Poisson integral of the same function but replaced by$0$on the right half of the circle. Then$0 < w < v$in the open disc. But$w(z)=0$on an arc of a circle near$1$, so by the Symmetry Principle,$\partial w/\partial r \neq 0$at the point$1$. As$w$is positive and$w(1)=0$, this derivative is negative. So$v(r)>w(r)>c(1-r)$for some$c>0$. 1 The limit is not zero; it does not exist. WLOG let$a=1$. Let$-u(z)$be your Poisson integral, (in the numerator of your formula) it is a negative harmonic function in the disc, continuous in the closed disc,$u(1)=0$, and negative at every other point of the circle. Let$M(r)=max_{|z|=r}u(z),\; 0\leq r\leq 1$. This is a strictly negative, increasing function on$(0,1)$, and$M(1)=1$. It is known that$M(r)$is convex with respect to the logarithm, that is $$r\frac{dM(r)}{dr}$$ is increasing. This is called Hadamard's Three Circles Theorem. Thus$M'(1)>0$. Now as$u(re^{i\theta})$is an even function of$\theta$decreasing on$(0,\pi)$, we conclude that$M(r)=u(r)$. This means that there exists a sequence$z_k\to 1$,such that$z_k<1$, and $$\frac{u(1)-u(z_k)}{1-z_k}\; \quad\to c>0.$$ So on this sequence your limit $$\frac{-u(z_k)}{|1-z_k|}\quad =c.$$ On the other hand it is clear that your limit is zero on sequences which tend to$1$"tangentially$, that is very close to the circle. So the limit does not exist.