Limit of arctan(1/(x^2+y^2)) at the Origin

The paths climb toward the same ceiling

Along a line through the origin $x^2+y^2$ is a constant times $t^2$, so the quantity inside the arctangent grows like a constant over $t^2$.

The values rise toward $\frac{\pi}{2}$ and never reach it, because $\arctan$ is strictly below $\frac{\pi}{2}$ at every real input. A limit does not have to be attained, and often is not.

An inside limit and an outside limit

Let $u = x^2 + y^2$. Away from the origin $u > 0$, and $u \to 0^+$ exactly when $(x,y) \to (0,0)$, so the inner expression $\frac{1}{u}$ runs to $+\infty$.

The composition step is legitimate because $\arctan$ is continuous on the whole real line and has the horizontal asymptote $\frac{\pi}{2}$ as its argument grows. Whatever route the point takes, the inner quantity ends up large and positive, and the outer function converts large and positive into something near $\frac{\pi}{2}$.

In polar coordinates $f = \arctan\left(\frac{1}{r^2}\right)$, with no $\theta$ at all. The function is constant on every circle around the origin, so direction cannot possibly matter and the path tests were never going to disagree.

The mistake: taking the inside limit and stopping

A common slip is to compute $\lim\left(x^2+y^2\right) = 0$, write $\arctan\left(\frac{1}{0}\right)$, and then either declare the limit undefined or write $\arctan(\infty)$ without saying what that means. $\frac{1}{0}$ is not a number. The honest statement is one sided: $u \to 0^+$ forces $\frac{1}{u} \to +\infty$, and that is what pins the answer to $\frac{\pi}{2}$.

The one sided part carries the whole argument. Replace the inside by $x^2 - y^2$ and the quantity approaches $0$ from both sides: $\frac{1}{x^2-y^2}$ runs to $+\infty$ along $y = 0$ and to $-\infty$ along $x = 0$.

Two paths, two different values, so that variant has no limit. The difference between the two problems is nothing more than a sign in the denominator, which is why the sign is worth stating out loud.