Consider the following diagram with stated conditions,
Using the ML lemma, it is possible to show that $$\int_{\Gamma_1} f(z) \ dz\rightarrow 0\ \ \text{as} \ \ R\rightarrow\infty, \ \ \text{as} \ \ \left|\int_{\Gamma_1} f(z) \ dz\right|\leq\frac{2\pi}{R}.$$
My question is, why can we not use a similar computation to find $$\lim_{r\to 0}\int_{\Gamma_3} f(z) \ dz?$$
For $\Gamma_3$, we can write \begin{align*} \lim_{r\to0}\int_{\Gamma_3}f(z)\,\mathrm{d}z&=\lim_{r\to0}\int_\pi^0 \frac{1-e^{ire^{i\theta}}}{r^2e^{2i\theta}}\,rie^{i\theta}\,\mathrm{d}\theta\\ &= \lim_{r\to0}\;i\int_0^\pi\frac{e^{ire^{i\theta}}-1}{re^{i\theta}}\,\mathrm{d}\theta\\ &= \lim_{r\to0}\;i\int_0^\pi \frac{\left(1+ire^{i\theta}-\frac{1}{2}r^2e^{2i\theta}+\cdots\right)-1}{re^{i\theta}}\,\mathrm{d}\theta\\ &= \lim_{r\to0}\,i\int_0^\pi\left(i-\frac{1}{2}re^{i\theta}-\frac{i}{6}r^2e^{2i\theta}+\cdots\right)\,\mathrm{d}\theta\\ &= \,-1\int_0^\pi1\,\mathrm{d}\theta\\ &=-\pi \end{align*}