Infinite limit of a complex function Question
$$ \lim_{x \to 0}\frac{1}{x^2 + i x} $$
The related info (following solution and plot taken from Wolfram Alpha)
$$
\lim_{x \to 0^-}\frac{1}{x^2 + i x} = i \infty
$$
$$
\lim_{x \to 0^+}\frac{1}{x^2 + i x} = (-i) \infty
$$

Why is the real part being ignored here when in plot it can be clearly seen that the real part is converging to 1 and imaginary part is diverging?
In my opinion the limit should be either $1-(\infty)i$  or  $1+(\infty)i$ depending on the side from which we approach.
If not can you clear my doubt how can we neglect the real part present here?
 A: Your opinion seems correct in one sense. I think the reason for the ambiguity is that in complex analysis, people usually only talk about one complex infinity (denoted just $\infty$ or $\hat\infty$). Here is the reasoning: multiplying your expression by the complex conjugate of the denominator, we get
$$\frac1{x^2+ix}=\frac{x^2-ix}{x^4+x^2}=\frac1{x^2+1}-\frac i{x^3+x}.$$
From here one easily computes the limits of the real and imaginary parts. It is certainly correct to say the real part approaches $1$ and the imaginary part approaches $\mp\infty$ as $x\to0^{\pm}$.
However less common is another notion called directed infinity which asks in which direction in the complex plane the convergence occurs. This would be something in the form $z\infty$, meaning infinity in the direction of $\arg(z)$. This must be what Wolfram Alpha is telling you, since asymptotically $1\pm i\infty$ is parallel to the ray starting at $0$ passing through $\pm i$.
A: Ok. The explanation of Funktorality is correct. If u consider the function it is a function from Reals to Complex. Now if u see the definition of limit is that as $x$ tends to $0$ $|f(x)-l|<\epsilon$ that mod is norm on the complex plane if such $l$ exist then it has a limit $l$. But certainly $l$ must belongs to complex. Now if u break the norm then real part and imaginary part must converge together to exist the limit.
