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!
Best Answer
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.