Why does $\lim_{z \to c} f(z)(z-c)=0$ imply that $\int\limits_{\partial B}f(\zeta)d\zeta=0$

I am using the book Theory of Complex Functions by Remmert. In chapter 7 there is a corollary that says for a function $f:D\to\mathbb{C}$, $D$ domain:
If $f$ is bounded around $z$, then $\int\limits_{\partial B}f(\zeta)d\zeta=0$. Where $B=B_{r}(c)$ and $\overline B\subset D$.

Then there is some sort of remark that says:
The integral already vanishes if f satisfies

$\lim_{z \to c} f(z)(z-c)=0$

I don't understand why this is true. A few pages later he talks about Riemann's Theorem about removable singularities. There he states the condition $\lim_{z \to c} f(z)(z-c)=0$ but uses things in the proof that are not "available" to me yet. I think it is possible to proof his remark using Goursat's, Morera's and Cauchy's Integral Theorem but Morera's Theorem is also not at my disposal yet.Furthermore there is version of Goursat's Theorem that makes weaker assumptions. This version is I think necessary to proof the claim.
What am I missing? How do you proof the claim only using the things that have been proven so far? Thank you very much in advance for your help!

• it says that if $f(z)$ is holomorphic/analytic on $0 < |z| < a$ and $\lim_{z \to 0} f(z) z = 0$ then for $0 < r < a$ : $\ \int_{|z| = r} f(z) dz = \lim_{ \epsilon \to 0} \int_{|z| = \epsilon} f(z) dz = 0$ (since $|\int_{|z| = \epsilon} f(z) dz| < \int_{|z| = \epsilon} |f(z)| d|z| = \epsilon \max_{|z| = \epsilon} |f(z)| = \max_{|z| = \epsilon} |z f(z)|$) – reuns Jul 8 '16 at 10:32
• do you mean the corollary or the remark? – Hans Jul 8 '16 at 10:34
• And when writing $\int_{|z| = r} f(z) dz = \lim_{\epsilon \to 0} \int_{|z| = \epsilon} f(z) dz$ I'm using the Cauchy integral theorem – reuns Jul 8 '16 at 10:35
• And for proving $f(z)$ is holomorphic/analytic at $z = 0$, I prefer to prove that $z=0$ is a pole or an essential singularity, and that in both case it would be contradicting $\lim_{ z \to 0} z f(z) =0$ (the essential singularity part requiring the Liouville theorem : the non-existence of a bounded entire function) – reuns Jul 8 '16 at 10:39
• Unfortunately Liouville has not been proven yet. – Hans Jul 8 '16 at 10:53