# The limit of a function from different directions: rotate the function and take limit

Let $$\{(x,y)\}_{x\in\mathbb R}$$ be a continuous curve; what will be implied if it is given that the any rotated version of that curve have a limit?

i.e. let $$(s,t)=A(x,y)$$ be a rotation around the origin, $$\lim_{s\to+\infty}(s,t)$$ exists or tends to infinity.

We say the mapping $$f$$ has a limit if the limit exists or tend to $$\infty$$. i.e. $$\lim_{x\to +\infty}f(x)=K\subset [-\infty,+\infty]$$ where $$K$$ is a finte set, $$f$$ is a correspondence.

First Case: the limit of a curve exists, and, the limit of the rotated curves also always exist. For example, the limit of the curve $$\lim_{x\to+\infty}(x,x)=(+\infty,+\infty)$$ exists, and the limit of $$\{(x,ax)\}_{x\in\mathbb R}$$ also exist for any $$a$$. In fact, the limit of any polynomial curve $$\{(x,f(x))\}_{x\in\mathbb R}$$ (where $$f(x)=\sum_n a_nx^n)$$ exist and so do their rotated point sets. We call that the curve "has a limit in every direction".

Second Case: Sometimes, the limit of a curve does not exist, but if rotate the curve, then the limit might exist. For example, the curve $$\{(x,\sin(x))\}_{x\in\mathbb R}$$ does not have a limit, but rotated curve $$\{(s,t)\}_{s\in\mathbb R}$$, where $$(s,t)=(\frac{1}{\sqrt 2}(x-\sin(x)),\frac{1}{\sqrt 2}(x+\sin(x))),$$ does have a limit: $$\lim_{s\to +\infty}(s,t)=(+\infty,+\infty)$$

Question: if we know that the curve {(x,y)} has limit in every direction, what can be implied from this kind of convergence? Does that mean that the function is asymptotically polynomial? Or, asymptotically not a periodic function?