What is the value of $\lim_{x \to 0} \tan^{-1}(x/m)$ if $m < 0$

Question

I have a quote from my book that says:

$\lim_{x \to 0} \tan^{-1}\left(\frac {x}{m}\right)=0$ if $m > 0$ or $\pi$ if $m < 0$

Added context from the book:

The goal is to find the value of $\displaystyle\int_0^{\infty}\frac{ \sin(mt)}{t}\mathrm{d}t$

So the above integral is written as the laplace transform:

$$\int_0^{\infty}e^{-st}\frac{ \sin(mt)}{t}\mathrm{d}t=\frac{\pi}{2}-\tan^{-1}\left(\frac {s}{m}\right)$$

And then it states the above (my question) and gets two different results for $\pm m$.

My thoughts:

It's pretty much obvious for the value to become zero for $m > 0$ but I am a bit confused for $m < 0$.

Even if I plot a graph(below) of $\tan^{-1}(-x)$. I can still see that as $x \to 0$ the value of $\tan^{-1}(-x) \to 0$

I mean if $x$ was a number or even $\infty$ it's understandable that it can be negative of that number or $-\infty$ but here $x \to 0$ so there nothing negative about $0$.

Can someone please let me know if I am wrong somewhere or my book is wrong?

Answer 1

Your book is wrong. The standard definition of the arc tangent is an odd function and $$\lim_{x\to0}\arctan x=0.$$

By the way,

$$\lim_{x\to 0^\pm}\arctan\frac xm=\lim_{x\to 0^\mp}\arctan\frac x{(- m)}$$ (limits to $\pm$ and $\mp$) and the existence of the limit self-contradicts the claim of the book.

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If you consider that an implicit reference to the function $\arctan_2$ is made, you have, with the usual convention,

$$\lim_{x\to0}\arctan_2(x,m)=0$$ for positive $m$, and

$$\lim_{x\to0^+}\arctan_2(x,m)=\pi,$$

$$\lim_{x\to0^-}\arctan_2(x,m)=-\pi$$ for negative $m$, so that the plain limit does not exist.

Answer 2

As you know that $$\arctan (-x)=- \arctan x $$

Thus if $m<0$ we can write it as

$$\arctan(\tfrac x m)=-\arctan(\tfrac x {-m})=-\arctan(\tfrac x n)\hspace{15pt}n>0 $$

Thus $$\lim_{x \to 0}\arctan(\tfrac x m)=-\lim_{x \to 0}\arctan(\tfrac x {-m})=-\lim_{x \to 0}\arctan(\tfrac x n)=-0=0 $$