Here are two limits with different answers provided by Wolfram|Alpha:
$$\lim_{z\to \frac{2}{5}-\frac{i}{5}} \frac{2\left(z-\frac{2-i}{5}\right)}{(1-2i)z^2+(6i)z+(-2i-1)}=-\frac{i}{2}$$
$$\lim_{z\to \frac{2}{5}-\frac{i}{5}} \frac{2\left(z-\frac{2-i}{5}\right)}{\left(z-\frac{2-i}{5}\right)\left(z-(2-i)\right)}=-1-\frac{i}{2}$$
Aren't these the same limits because $(1-2i)z^2+(6i)z+(-2i-1)=\left(z-\frac{2-i}{5}\right)\left(z-(2-i)\right)$?
Here is what Wolfram|Alpha suggests.
The two denominators are not equal : notice that they don't have the same coefficient for $z^2$. Instead, you should have (according to the $3$rd link) the equality $$(1-2i)z^2+(6i)z+(-2i-1)=(1-2i)\left(z-\frac{2-i}{5}\right)\left(z-(2-i)\right),$$ so the two limits only differ from a factor $1-2i$. Then the results become coherent, as indeed, $$-\frac{i}{2}\left(1-2i\right)=-1-\frac{i}{2}.$$