Prove that the function $\mathbb{R} \to \mathbb{R}^2: t \mapsto \left(t, at^2 + bt +c\right)$ doesn’t have an asymptote.
Here what I thought so far:
First of all Im using this definition for a line $L$ to be an asymptote of a curve:
$$ \lim\limits_{t \to A} d(f(t),L) = 0 $$
Where $A$ can be $\pm \infty$ and $\lim\limits_{t \to A} \left|f(t)\right| = \infty$.
The first solution I came up with is more conceptual, and since I want to practice my formal proof writing, I want to prove it using a more formal (delta epsilon) fashion.
Here’s the first solution:
- By the definition of asymptote and the continuity of $f(t)$ and since $d(f(t), L) \geq \left|f(t) - L\right|$:
By contradiction, let’s assume the function does have an asymptote L. Then
$$ \lim\limits_{t \to A} \left|f(t) - L\right| = \left| \lim\limits_{t \to A} f(t) - L\right|= 0 \implies \lim\limits_{t \to A} f(t) = z \in L $$
Which is, the sequence in $f(t)$ converges to a point in the line. So now, since the point $z$ lies in the line, we can use the fact that their derivative vector also matches the lines direction. But using the Lagrange Integral error bound for the parabola we can check that the function and the line at $z$ at the direction of it’s derivative diverges, which is, the derivative should match $L$’s derivative through any point, as well as the function $f(t)$ should be arbitrarily close to $L$, but the integral error shows the opposite. A contradiction.
Although this first solutions seems to make sense, it seems to lack formalism since Im using a point within the line on the limit. So I though of another way:
- Since the function $f(t)$ has a well defined and simple expression, explicitly calculate the distance between $f(t)$ and its projection onto a generic parametric line $L(s) = p + vs $. This yields a quadratic expression that clearly does not converges to zero. So it shows no generic line can indefinite approach the parabola.
My questions are: are those 2 solutions reasonable or correct?
Second question is: is there a more formal delta epsilon way to solve it (possibly also more elegant?).
Thanks!
If $z$ is a point of $L$, then by definition $$d(f(t),L) \le |f(t)−z|$$ and not the other way around, and this invalidates your first proof.
The second approach works, but of course requires some calculations. Anyway, there are some characterizations that allows for a somewhat simpler proof. For instance, with your notations, a function has an asymptote (at infinity) iff $$ \exists \lim_{t\to\pm\infty} \frac{f(t)}{t} := m\in \mathbf{R}$$ and $$ \exists \lim_{t\to\pm\infty} f(t)-mt := q\in \mathbf{R}.$$ In this case the line $t\mapsto mt+q$ is the asymptote.
On the contrary in your case $ \lim_{t\to\pm\infty} \frac{f(t)}{t} $ is not finite, hence there is no asymptote.