Why is $\lim_{x\to\infty} \frac{1}{x}$ equal to $0$, when really the limit appears to be an infinitesimal quantity?
I am trying to understand why there is no distinction between 0 and an infinitesimal quantity in the context of limits.
On
The limit of an equation also affects the answer. An infinitesimal is a limit to 0, but since the lim symbol is already there, we can write it as 0.
On
If the limit appears to be an infinitesimal quantity, then you probably have one of the common fundamental misconceptions about limits are doing.
In particular, you are failing to conceive of the limit, but are instead thinking of the behavior of the function near the limit.
Misconceptions about $\pm \infty$ often muddle the issue here. For comparison, note that $\lim_{x \to 0} x$ is also exactly zero, not merely 'infinitesimal'. $\lim_{x \to \infty} \frac{1}{x}$ isn't fundamentally any different from $\lim_{x \to 0} x$, but sometimes the fact $\infty$ is involved sometimes gives people a mental block.
On
In the early days of calculus the idea of infinitesimals or "vanishingly small quantities" different from $0$ was used somewhat freely although those doing so knew that they had no rigorous justification for doing so.
At some point mathematicians such as Cauchy and Bolzano tried to treat the notion of limit with more rigor by taming the concept of "infinity" in their arguments.
But it was Weierstrass who finally put this into the form of a definition, the $\epsilon, \delta$ definition which we know today.
Thus mathematicians could speak plainly and say that
$$\lim_{x\to\infty}\frac{1}{x}=0$$
rather than trying to explain what they meant by "the smallest possible number greater than $0$" or other circumlocutions.
In modern times others tried to logically incorporate a notion of "infinitesimals" into calculus in what is called "non-standard analysis."
In standard real analysis/calculus, there are no infinitesimal quantities. Everything is formulated in terms of real numbers. What $\lim_{x\to \infty} f(x) = c$ means is that for all $\varepsilon > 0$ there exists $x_o\in \mathbb{R}$ such that whenever $x>x_0$, we have that $\vert f(x)-c\vert < \varepsilon$. In words, what this means is that if you pick and small number $\varepsilon$, you can find a number $x_0$ large enough such that at any number past $x_0$, $f(x)$ will be no greater than a distance $\varepsilon$ away from $c$. For example, if $f(x)=\frac{1}{x}$ and $c=0$, this is the case. Given $\varepsilon > 0$, we can let $x_0 = 1/\varepsilon$; then if $x>x_0$, we have $\vert f(x)-0\vert = 1/x < \varepsilon$. So, $1/x$ really does approach $0$, in that it gets arbitrarily close to $0$.