I had this as part of a question in an exam. And, I reasoned, even when it's arctan(1/0) (undefined), it is pi/2. And, so I said, domain belongs to all Real Numbers. Why isn't it this
[Math] Domain of arctan(1/x)
inversetrigonometry
Related Solutions
Remember that $\arcsin$ is supposed to be the inverse function of $\sin$ (or at least, of a "restricted sine function"; that's the choice of 'principal value').
The way that $\arcsin$ works is supposed to be: you plug in the value somebody else got out of the sine function, and $\arcsin$ will tell you what number was put into the sine function to get that value. It's like a "reverse telephone directory": you look up the phone number and find out the person it belongs to, instead of the usual way of looking up the person and finding their phone number.
But that means that the only things that you can put into the $\arcsin$ functions are real numbers that actually come out of the sine function (the only numbers you can look up in a reverse telephone directory are telephone numbers, so you can't look up "000-0000").
What are the numbers that can come out of the sine function? Every number between $-1$ and $1$ (inclusively), but only those numbers: the sine function will never give a result that is greater than $1$ or smaller than $-1$. That means that the numbers you can plug into $\arcsin$ are only the numbers that come out of the sine function: the numbers between $-1$ and $1$.
The same thing is true for $\arccos$: you can only put into $\arccos$ numbers that may come out of the cosine function, and the only numbers that may come out of the cosine function are the numbers on $[-1,1]$.
However when we come to the $\arctan$ function, things are different: what are the numbers that may come out of the tangent function? Every real number! Every real numbers is the tangent of some angle, so now we can put into $\arctan$ any real number, because any real number is, potentially (and in actuality) the result of applying the tangent function.
Note. Your final paragraph seems to be confusing the "trigonometric tangent function" with the tangent line to the graph of a function. The "trigonometric tangent function" is the function defined by $$\tan(x) = \frac{\sin(x)}{\cos(x)}.$$ The "tangent line to the graph of $f(x)$ at $x=a$" is a straight line that has certain properties (it goes through $(a,f(a))$, and is the straight line that "best approximates" the graph of $y=f(x)$ near the point $(a,f(a))$). The derivative, which is a key concept in calculus, is the slope of the tangent-line-to-the-graph-of-$f(x)$ (which is defined as a limit of a certain ratio), not to the trigonometric tangent function which is what $\arctan(x)$ is related to.
Right triangle with legs $1$ and $x$, just remembering the sum of all the angles of a triangle, is $\pi$. In case of a right triangle, $\alpha+\beta={\pi\over2}$ for any $x$.
Best Answer
$$\lim_{x\to0^+}\dfrac1x=+\infty\neq\lim_{x\to0^-}\dfrac1x=-\infty,\qquad\lim_{y\to+\infty}\arctan(y)=\dfrac\pi2\neq\lim_{y\to-\infty}\arctan(y)=-\dfrac\pi2$$ Since the left limit differs from the right limit, the limit does not exist. To exist, the two limits must be equal and finite.