Finding the asymptotes of the trajectory of a point parameterized by $(x,y)=\left(\frac{t}{1-t^2},\frac{t-2t^3}{1-t^2}\right)$ for $t\geq 0$

Question

Let $(x,y)$ be Cartesian coordinates in the plane and suppose a moving point has coordinates $$x=\dfrac{t}{1-t^2}, \quad y=\dfrac{t-2t^3}{1-t^2}$$ at time $t\ (t\geq 0)$. Describe the trajectory of the point and find asymptotes.

Solution: I sketched the trajectory of the point and it was not so difficult. Firstly we need to sketch $x(t)$ and $y(t)$ then we need to "combine" them in order to get the graph of $(x,y)$.

But I have some issues with finding asymptotes.

Definition: The line $c_0+c_1x$ is called an asymptote of the graph of the function $y=f(x)$ as $x\to - \infty$ (or $x\to +\infty$) if $f(x)-(c_0+c_1x)=o(1)$ as $x\to -\infty$ (or $x\to+\infty$).

Proposition: The line $y=kx+b$ is an oblique asymptote for the function $f(x)$ as $x\to +\infty$ if and only if $\lim \limits_{x\to +\infty}\dfrac{f(x)}{x}=k,\ k\in \mathbb{R}$ and $\lim \limits_{x\to +\infty} (f(x)-kx)=b, \ b\in \mathbb{R}$.

Since $\lim \limits_{t\to 1}\dfrac{y(t)}{x(t)}=-1$ and $\lim \limits_{t\to 1} ((y(t)+x(t))=2$, the line $y=-x+2$ is an asymptote for both ends of the trajectory, corresponding to $t$ approaching $1$. It is also clear that the line $x=0$ is a vertical asymptote for the portion of the trajectory corresponding to $t\to+\infty$.

This is an excerpt from Zorich's book.

Our parametric curve $(x(t),y(t))$ defines a function $f:\mathbb{R}\to \mathbb{R}$. Zorich is doing limit computation in terms of $t$. Can anyone explain in a rigorous way why it implies that $\lim \limits_{x\to +\infty}\dfrac{f(x)}{x}=-1$ and $\lim \limits_{x\to +\infty} (f(x)+x)=2$.

Would be very grateful for help!

Answer 1

As $y=(1-2t^2)x$ and $x \to +\infty$ when $t\to -1-0$ and $t\to 1-0$, then $\lim\limits_{x\to+\infty}\frac{y}{x}=\lim\limits_{t\to1-0}(1-2t^2)=-1$. For second $y+x=2t\to 2$, when $t\to 1$.

Addition. In left neighbourhood of $1$ function $x=x(t)$ is continuous and strictly monotonic, so exists $t=t(x)$ inverse function, so we can consider composition $y(t(x))$. For formal reasoning for limit of composition we can use following, true for one or several variables, theorem:

Suppose exists, finite or not, limits $\lim\limits_{x \to a}f(x)=b$ and $\lim\limits_{y \to b}F(y)$. If in some punctured(deleted) neighbourhood of $a$ holds $f(x) \ne b$, then in point $a$ exists limit of composition and holds $$\lim\limits_{x \to a}F(f(x))=\lim\limits_{y \to b}F(y)$$ proof of this theorem can be found in Kudryavtsev L.D., Course of mathematical analysis, volume 1, 1981, pages 108, 326.

In your case taking $F(t)=1-2t^2$, for composition we have $F(t(x))=1-2t^2(x)=\frac{y(t(x))}{x}$, so knowing limit for $F(t)$ gives limit for $F(t(x))$: $$\lim\limits_{x\to+\infty}\frac{y(t(x))}{x}=\lim\limits_{x\to+\infty}(1-2t^2(x))=\lim\limits_{t\to1-0}(1-2t^2)=-1$$

Answer 2

The trajectory is defined for $t \ge 0$ and $t \neq 1$. Let $I= [0,1) \cup(1, \infty)$.

$x$ is a strictly increasing continuous bijection $[0,1) \to [0,\infty)$ and a strictly increasing continuous bijection $(1,\infty) \to (-\infty,0)$. Hence $(x(t), y(t))$, $t \in I$ defines a function $f:\mathbb{R} \to \mathbb{R}$.

It is straightforward to check that the maps $x:[0,1) \to [0,\infty)$ and $x:(1,\infty) \to (-\infty,0)$ are actually homeomorphisms.

It is straightforward to check that $\lim_{s \to \infty} \phi(s) = L$ iff $\lim_{t \uparrow 1} \phi(x(t)) = L$.

Then we have $\lim_{s \to \infty} {f(s) \over s} = \lim_{t \uparrow 1} {f(x(t)) \over x(t)} = \lim_{t \uparrow 1} {y(t) \over x(t)} = \lim_{t \uparrow 1} (1-2t^2) =-1$, and similarly, $\lim_{s \to \infty} {f(s) + s} = \lim_{t \uparrow 1} {f(x(t)) + x(t)} = \lim_{t \uparrow 1} {y(t)+ x(t)} = \lim_{t \uparrow 1} 2t =2$.

Note that it is also true that $\lim_{t \to 1} {y(t) \over x(t)} = \lim_{t \to1} (1-2t^2) =-1$ and $\lim_{t \to 1} {y(t)+ x(t)} = \lim_{t \to 1} 2t =2$, and using the $x:(1,\infty) \to (-\infty,0)$ portion of the curve, we can show in a similar manner that $\lim_{s \to -\infty} {f(s) \over s} = -1$ and $\lim_{s \to -\infty} {f(s) + s} = 2$.

Elaboration

Take the case where $L$ is finite for example. Also, note that I am dealing with $x:[0,1) \to [0,\infty)$ here. Note that $x$ is strictly increasing on this domain:

Suppose for all $\epsilon>0$ there is some $S$ such that if $s > S$ then $|\phi(s)-L| < \epsilon$. Now let $t_0= x^{-1}(S), \delta = 1-t_0$ and note that if $0< |1-t| < \delta$ (and implicitly we have $t \in [0,1)$) we have $|\phi(x(t))-L| < \epsilon$.

The other direction is similar, suppose we have some $\delta>0$ such that if $0< |1-t| < \delta$ (and implicitly $t \in [0,1)$) then $|\phi(x(t))-L| < \epsilon$. We can assume that $\delta<1$. Let $S=x(1-\delta)$, then if $s > S$ there is some $t \in (1-\delta,1)$ such that $x(t)=s$ and so $|\phi(s)-L| = |\phi(x(t))-L| < \epsilon$.

The case for infinite $L$ is similar.

Answer 3

hint

With cartesian equation $ y=f(x) $, we look for oblic asymptotes Only when $$\lim_{|x|\to\infty}f(x)=\pm\infty$$ we look for $$\lim_{x\to\infty}\frac{f(x)}{x}=a$$ and $$\lim_{x\to\infty}(f(x)-ax)=b$$

This cartesian representation can be seen as a special parametric one if we put $$x(t)=t \;\text{ and } \;y(t)=f(t).$$

(In parametric representation with the variable $ t $, You need $ t_0 $, such that both $ x(t) $ and $ y(t) $ go to infinity when $ t $ goes to $ t_0$. then $$\lim_{x\to \infty} \frac{f(x)}{x}$$ becomes $$\lim_{t\to t_0}\frac{y(t)}{x(t)}$$

and $$\lim_{x\to \infty}(f(x)-ax)$$ is replaced by $$\lim_{t\to t_0}(y(t)-ax(t))$$

In your case $ t_0\in\{-1,1\}$ and knowing that $ t\ge 0$, we have only the case $t_0=1$.