Does $e^{-x} $ "intersects" x axis at "infinity"? Does $e^{-x} $ "intersects" x-axis at "infinity" ?

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*From what I understand it gets closer and closer so it must intersect and so it  intersect at "infinity" where infinity means what it means in "all other " problems.

*If there is no answer then someone should point out why is it and what are the reasons for that. And possible directions one should proceed to make the  answer defined.
Just like $ \sqrt{-1} $ did not make sense but is useful . Could this be the case here. My purpose is just to gain a better understanding of problem.

 A: What does it mean to say the curve "meets" or even "intersects" the $x$-axis at "infinity"? What do you mean here by "infinity"? In a literal sense, $y = e^{-x}$ never meets the $x$-axis. To do so, you'd need an $x$ for which $e^{-x} = 0$ and there is no such value. But some are fond of using "meeting" (asymptotes and parallel lines) or "focussing" (lens theory) "at infinity" in situations where literal meeting or focussing never actually occur. This is simply a quirk of terminology. One shouldn't infer too much from these expressions in my view.
A: $\newcommand{\Reals}{\mathbf{R}}$To flesh out the point made in the comments, in the real plane $\Reals^{2}$ we define the positive $x$-axis to be the set of points $(x, 0)$ with $x > 0$ real, and define the positive $y$-axis to be the set of points $(0, y)$ with $y > 0$ real. Do these curves intersect at $(0, 0)$? No they do not, because the origin lies on neither curve.
In this example, we might have an overwhelming urge to "extend" the positive axes, either to the non-negative axes, or perhaps to the Cartesian axes. In either of those cases, the axes do meet at the origin (and as it happens nowhere else).
To obtain this "yes" answer, we had to ask a different (but related) question involving two additional ingredients:

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*A larger "completed space", namely a set with some additional structure containing our curves, here the Cartesian plane with the Euclidean distance function and its induced notions of limits and continuity.

*A way to "extend" our curves in the completed space.

In the positive-axes analogy, the completed space was implicitly present in the phrasing of the question (but see also the final note below).
To give a second analogy, a bit closer to your question because the choice of "completion" is not obvious, "Do parallel lines in the plane meet?". The usual definition of parallel for lines in the plane is non-meeting, so again the literal answer is "no". On the other hand, geometers, artists, and others will often say "parallel lines meet at infinity"!

Now to your question, "Does the $x$-axis intersect the graph $y = e^{-x}$?": We can accept that the answer is "no" because the graph $y = e^{-x}$ in the Cartesian plane (the implicit universe for such a graph) has no point with $y = 0$.
Alternatively, we can look for a different but related question along the conceptual lines above. As with the positive axes, or with parallel lines, we need a "completion" and some way of extending our curves.
There are multiple ways of completing the Cartesian plane $\Reals^{2}$ by adding "points at infinity". Of the five that come immediately to mind, here are three:

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*Introduce "signed infinities" into the reals, obtaining $\overline{\Reals} = \Reals \cup\{-\infty, +\infty\}$, and form the extended Cartesian plane $\overline{\Reals}^{2}$. (This is not a common construction, but is mathematically consistent and easy to draw.) In this setting, there is a square of "points at infinity", and these form the boundary of the extended plane. The diagram below depicts the extended plane by mapping $(x, y)$ to $(\tanh\frac{x}{4}, \tanh\frac{y}{4})$. The graph $y = e^{-x}$ is shown, and the prospective point of intersection is the open circle on the right.

*Introduce one "point at infinity" for every Euclidean direction, obtaining the real projective plane. (Projective plane geometry is a large, well-trodden realm.)

*Introduce one "point at infinity" period. The result is seen to be a sphere via stereographic projection, a fact amply explained elsewhere on site. (This is the usual construction in complex analysis, and is also well-trodden.)


In each of these three settings, the $x$-axis in the real plane $\Reals^{2}$ has a unique "limit point" as $x$ grows without bound; the graph $y = e^{-x}$ also has a unique limit point as $x$ grows without bound; and these limit points coincide.
In this sense, we might say, "Yes, the $x$-axis meets the graph $y = e^{-x}$ at infinity."
The crucial points are that we have asked a different question about a more-or-less natural geometric reframing, using the same language and wording.

The contrast of these three "completions" of the Cartesian plane should impress the need for care. The question about the $x$-axis and the graph $y = e^{-x}$ turned out not to depend much on our choice of completion. If we ask about parallel lines, however, the choice is crucial:

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*In the first model, any two parallel lines of positive slope meet at two points, $(+\infty, +\infty)$ and $(-\infty, -\infty)$; similarly, parallel lines of negative slope meet at two points, $(+\infty, \infty)$ and $(-\infty, +\infty)$; horizontal lines meet at two points, $(+\infty, 0)$ and $(-\infty, 0)$; and vertical lines meet at two points, $(0, +\infty)$ and $(0, -\infty)$.

*In the second model, which is inspired by perspective projection, there is one point at infinity for each family of parallel lines (think of railroad tracks), and all lines from such a family meet at that point.

*In the third model, all lines pass through the single point at infinity.

One final note: In each of our three completions we found that the $x$-axis does intersect the graph $y = e^{-x}$. Does this (doesn't this) mean the answer is yes, they intersect?
Briefly, no it doesn't. To illustrate, let's return to the analogy of the positive coordinate axes, viewed as subsets of the punctured plane, the Cartesian plane with the origin removed. There is a useful way to "complete" the plane at the origin by adding, rather than one point, one point for each line through the origin, a little like the way we added points at infinity to get the projective plane. (Mathematicians call the resulting space the blow-up of the plane at the origin.) In this setting, the positive axes do extend, but the extensions do not intersect.
A: See it is not "mathematically good" to say that the graph of $e^{-x}$ intersects the x-axis at infinity. It is just a crude way of saying that limiting value of the function is $0$ as value of x increases i.e. it gets closer and closer to zero but never reaches it as x increases. No matter how large the value of x, there will be some distance between it's graph and x-axis .
Why I said not mathematically good because intersection of graph with a line is defined by a common point where this intersection occurs, since there is no common point for intersection of graph of $e^{-x}$ and x-axis (as there's no real solution to the equation  $e^{-x}=0$ ), we can't define their intersection.
