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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

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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.

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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 $x\rightarrow0^-$

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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?

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