# Find the limit of $\arctan(y/x)$

I need to calculate the limit of function $$\arctan(y/x)$$ for the arguments $$(x,y)\to(0,1)$$.

Earlier I thought it's a bit easy but I just cannot get any answer, Instinctively it may look like limit is $$\pi/2$$ but one could also argue about two paths yielding different limits.

Kindly help!!

Thanks & regards

As $$(x,y) \to (0,1)$$ through positive values of $$x$$ the ratio $$\frac y x \to \infty$$ and it tends to $$-\infty$$ as $$(x,y) \to (0,1)$$ through negative values of $$x$$. Hence $$arctan(\frac y x)$$ tends to $$+\frac {\pi} 2$$ as $$(x,y) \to (0,1)$$ through positive values of $$x$$ and to $$-\frac {\pi} 2$$ as $$(x,y) \to (0,1)$$ through negative values of $$x$$. So the limit does not exist.
Does not exist because in our case $$\arctan{\frac{y}{x}}\rightarrow\frac{\pi}{2}$$ for $$x\rightarrow0^+$$ and $$\arctan{\frac{y}{x}}\rightarrow-\frac{\pi}{2}$$ for $$x\rightarrow0^-$$
Hint. Compare the limits along the line $$y=1$$ from the left and from the right with respect to $$x=0$$: $$\lim_{x\to 0^+} \arctan(1/x)\quad\text{and}\quad \lim_{x\to 0^-} \arctan(1/x).$$ Do you get the same result? What may we conclude?