There is a function given:
$$f(z) \in C\big(D'(a, r) \big),$$
where $D'(a,r) = D(a,r) \setminus \{a\}$.
We also know that:
$$\lim_{z \to a} \big(f(z)(z-a) \big) = A.$$
With the knowledge above we are to prove the following statement:
$$\lim_{r \to 0} \int \limits_{\gamma _r} f(z) \mbox{d}z = iA(\alpha-\beta),$$
where $\gamma_{r} =a + re^{it}, t\in [\alpha, \beta]$.
My attempt
Unfortunately, because we don't know if $f \in H\big(D'(a, r) \big)$ we can't use residues to calculate this integral.
I though of computing it in a standard way:
$$\int \limits_{\gamma _r} f(z) \mbox{d}z = \int \limits_{\alpha}^{\beta} f\big(a + re^{it} \big) ire^{it} \mbox{d}t = ir \int \limits_{\alpha}^{\beta} f \big(a + re^{it} \big) e^{it} \mbox{d}t.$$
I don't really know where to go from here.
This should work out nicely. Just write
$$f(z) = \frac{A}{z-a} + \frac{\epsilon(z)}{z-a},$$
where $\lim_{z\to a} \epsilon(z) =0.$ Put that into the second integral you wrote in your question and good things will happen.