How to solve this complex limit $ \lim_{N \to \infty} \arg \left( \frac{1}{Nj+1}\right) $

Question

How to solve this limit $$ \lim_{N \to \infty} \arg \left( \frac{1}{Nj+1}\right) $$

The prove that I have seen is using that the $\arg$ of a quotient is the diference beetween the num and den i,e:

$$ \arg \left( \frac{1}{Nj+1}\right)=\arg \left( 1\right)-\arg \left( Nj+1\right)= 0-\arg \left( Nj+1\right)=-\arg \left( Nj+1\right) $$

Then how $A \to \infty$ then I can ignore the $1$(This do not convice me very much) and I get $$ = -\arg \left( Nj\right) $$ So $\lim_{N \to \infty}-\arg \left( Nj\right)=\lim_{N \to \infty}-\arg \left( j\right)= -90^{deg}$

But in this last line I can really forget de N?, and why?

Answer 1

Well, assuming that $\text{j}^2=-1$:

$$0^\circ=0\le\arg\left(\frac{1}{1+\text{N}\text{j}}\right)=\arg\left(1\right)-\arg\left(1+\text{N}\text{j}\right)=0-\arctan\left(\frac{\text{N}}{1}\right)\le2\pi=360^\circ$$

And:

$$\lim_{\text{N}\to\infty}\arctan\left(\text{N}\right)=\frac{\pi}{2}=90^\circ$$

Answer 2

We could first simplify :

$$\frac1{Ni+1}=\frac{1-Ni}{1+N^2}\implies \arg\frac1{1+Ni}=\arctan(-N)\xrightarrow[n\to\infty]{}-\frac\pi2$$

In degrees, the above would be $\;-90^\circ\;$