Proof of Goursat's integral lemma when a function has a removable singularity

Question

During the course of complex analysis, our professor stated Goursat's integral lemma as follows: "let $f$ be continuous on $D$ and be analytic on $D−z_0$, then the integral of $f$ on the perimeter of a rectangle is $0$". The proof she gave us is the following: $$|I|=|\int_{\partial R}f(z)dz|\le\sum_{i=1}^4|\int_{\partial R_i}f(z)dz|\le4|\int_{\partial R_1}f(z)dz|$$ I is the integral we want to show is $0$, $\partial$R is the perimeter of the rectangle: then we divide $R$ into $4$ smaller rectangles $R_i$. In the last step, $R_1$ is the rectangle on which the integral has its maximum value. Repeating this procedure many times (dividing $R_1$ into more smaller rectangles) we reach this point: $$|I|\le4^k|\int_{\partial R_k}f(z)dz|$$ Now we consider $z_0\in R_k$: we know $f$ is analytic in $z_0$, so: $$\forall\epsilon>0\space\exists\delta>0: \text{ if } \space|z-z_0|<\delta\space \text{ then }\space |f(z)-f(z_0)-f'(z_0)(z-z_0)|<\epsilon|z-z_0|$$ If I split $R$ in enough rectangles (if $k$ is big enough) I can say that $$|z-z_0|<\delta \space\forall z\in R_k$$ I can choose $\delta=L/2^k$ where $L$ is the measure of the perimeter of $R$ and $L/2^k$ is the perimeter of $R_k$. So $|z-z_0|<L/2^k$ and: $$|f(z)-f(z_0)-f'(z_0)(z-z_0)|<\epsilon L/2^k$$ Now I can write: $$|\int_{\partial R_k}f(z)dz|=|\int_{\partial R_k}(f(z)-f(z_0)-f'(z_0)(z-z_0))dz|$$ This is because $f(z_0)$ and $f'(z_0)(z-z_0)$ are continuous with their primitive function continuous too, so their integral is $0$ (we already proved this before this theorem). So: $$|\int_{\partial R_k}f(z)dz|=|\int_{\partial R_k}(f(z)-f(z_0)-f'(z_0)(z-z_0))dz|\le \int_{\partial R_k}|f(z)-f(z_0)-f'(z_0)(z-z_0)|dz\le \epsilon L/2^k\int_{\partial R_k}dz=\epsilon L^2/4^k$$ In conclusion, $$|I|=|\int_{\partial R}f(z)dz|\le4^k|\int_{\partial R_k}f(z)dz|\le 4^k \epsilon L^2/4^k\le \epsilon L^2$$ Since we can take epsilon as small as we want, then I must be $0$. This is the proof if there isn't any removable singularity. In case there is a singularity, if it is in a smaller rectangle which is not the one with the biggest value of the integral, then the proof doesn't change. If it is exactly in the $R_k$ rectangle we consider in the end, of course we can't say $f$ is analytic there, so, instead of this step: $$|\int_{\partial R_k}f(z)dz|=|\int_{\partial R_k}(f(z)-f(z_0)-f'(z_0)(z-z0))dz|\le \int_{\partial R_k}|f(z)-f(z_0)-f'(z_0)(z-z_0)|dz\le \epsilon L/2^k\int_{\partial R_k}dz=\epsilon L^2/4^k$$ we use the Darboux inequality, saying that: $$|\int_{\partial R_k}f(z)dz|\le \max(|f(z)| \space \text{ on } \space R_k)\cdot\partial R_k$$ Since $\partial R_k$ tends to $0$ when $k$ is big, the integral tends to $0$ too. After this I asked my professor why then we couldn't use this last simple proof even if there was not a removable singularity, since it seemed much easier than everything done before. She said it was not possible, because $$|I|\le4^k|\int_{\partial R_k}f(z)dz|$$ so we can't just show that $\int_{\partial R_k}f(z)dz$ tends to $0$: we have to make sure it does "faster" than $4^k$. Which makes a lot of sense, but, if so, what we showed at the end for the case with a removable singularity is useless, since we only know that $\int_{\partial R_k}f(z)dz$ tends to $0$. Could you help me understand?