For a few days I am looking at proofs for Cauchy's integral formula, but something still does not add up to me: $$\int\limits_C\dfrac{f(z)}{z-z_0}dz=\int\limits_C\dfrac{f(z_0)}{z-z_0}dz+\int\limits_C\dfrac{f(z)-f(z_0)}{z-z_0}dz$$ Then one uses the $\varepsilon,\delta$ definition to proof that $$\left|\int\limits_{C_r}\dfrac{f(z)-f(z_0)}{z-z_0}dz\right|\le\cdots<\varepsilon$$ But all this only proves that $$\lim\limits_{r\to0}\int\limits_C\dfrac{f(z)}{z-z_0}dz=\lim\limits_{r\to0}\left[\int\limits_{C_r}\dfrac{f(z_0)}{z-z_0}dz+\int\limits_{C_r}\dfrac{f(z)-f(z_0)}{z-z_0}dz\right]$$ meaning: I do not see how the equations of limits imply that the integrals themselves are actually equal.
Where is my mistake?
The paths $C$ and $C_r$ are homotopic in $\mathbb C\setminus\{z_0\}$ and therefore$$\int_C\frac{f(z)-f(z_0)}{z-z_0}\,\mathrm dz=\int_{C_r}\frac{f(z)-f(z_0)}{z-z_0}\,\mathrm dz.$$Bet the integral to the right of the $=$ sign is arbitrarily small. Therefore, it is equal to $0$. So, the integral to the left of the $=$ sign is also equal to $0$ and so\begin{align}\int_C\frac{f(z)}{z-z_0}\,\mathrm dz&=\int_C\frac{f(z_0)}{z-z_0}\,\mathrm dz+\int_C\frac{f(z)-f(z_0)}{z-z_0}\,\mathrm dz\\&=\int_C\frac{f(z_0)}{z-z_0}\,\mathrm dz\\&=2\pi if(z_0).\end{align}