Confusion about limits, specifically the definition of a limit

Question

$$\lim_{x\to0}f(x)$$ The most common way to read this is "limit of f(x) as x approaches 0". But the thing is, I find that there is a difference between these two limits for example $$\lim_{x\to\infty}\frac{x^2-1}{x^2+1} = 1$$ $$\lim_{x\to0}\frac{x}{5} = 0$$ Looking at the first one, technically speaking, the value won't actually ever hit 1. But it's converging to it. But with the second one, it does become 0, even though it's correct to say "as x approaches 0, $\frac{x}{5}$ approaches 0", it's also correct to say "as x approaches 0, $\frac{x}{5}$ will eventually equal 0". The latter statement doesn't sound so correct when you try to apply it with the first limit.

This kind of confused me, and I was unable to find answers online, so for a while I made a deduction myself and kept in mind that "yes, $\frac{x^2-1}{x^2+1}$ won't actually ever equal 1, but it will be infinitesmally close to it, to the point that this infinitesmally small distance doesn't matter, thus concluding that from a larger scale it's equal to 1".

My question: is this interpretation correct?

Graphing the function of the first limit shows this even better

Edit: Sorry I edited the example because I realized the infinity limit is a bad a example!

Answer 1

No, by definition, a limit never "hits" the target value of the argument. To evaluate a limit, you must not use the value of $f(x_0)$.

Intuitively, a limit is the value we should reach for the given value of the argument, by comparing to the values in the neighborhood. It is not the value we do reach by using the function value.

When the two values are defined and coincide (I mean $\lim_{x\to x_0}f(x)$ and $f(x_0)$), we say that the function is continuous. But they don't have to.

Answer 2

I guess the major issue you are having is that you are using an intuitive understanding of limit, instead of a rigorous definition. The rigorous definition for limit at a finite point $x_{0}$ in calculus is as follows: $$\lim_{x\to x_{0}}f\left(x\right) = s \longleftrightarrow \forall \epsilon > 0, \exists \delta > 0, \forall x, \lvert x-x_{0}\rvert < \delta \longrightarrow \lvert f\left(x\right) - s \rvert < \epsilon.$$ Similarly, the definition for limit at infinity is as follows: $$\lim_{x\to\infty}f\left(x\right) = s \longleftrightarrow \forall \epsilon > 0, \exists y \in \mathbb{R}, \forall x, x > y \longrightarrow \lvert f\left(x\right) - s\rvert < \epsilon.$$

I am now using the definition to prove that $\lim_{x\to 0}\frac{x}{5} = 0$, and leave the rest to you. Given an arbitrary $\epsilon$, as long as $\lvert x-0\rvert < 5\epsilon$, we have $$\lvert\frac{x}{5}-0\rvert = \frac{\lvert x\rvert}{5} < \frac{5\epsilon}{5} = \epsilon.$$ According to the definition, we take $\delta = 5\epsilon$ for each $\epsilon$, and we have $\lim_{x\to 0}\frac{x}{5} = 0$.

Intuitive understanding is good for introduction of a concept. However, when it comes to learning rigorous math, understanding and using formal definition is crucial. Sometimes, a function is not defined at a point, but limit at that point exists. For instance, we have

$$\lim_{x\to 0}\frac{\sin{x}}{x} = 1.$$ The function $\frac{\sin{x}}{x}$ is not defined at $x = 0$, but according to the definition, the limit exists at that point. Thus, you should follow strictly the formal definition of limit, instead of intuitive understanding. Math is extracted from intuition, but it is not intuition itself. It is a set of deduction rules and axioms that incorporate intuition while show something farther.

Answer 3

Limits tell you about the behaviour of a function when you get close to a point, but not actually when you get there. For example, $$ \lim_{x \to 1}\frac{x^2-1}{x-1}=2 $$ but this does not mean that $0/0=2$. This is one of many examples of where you can't just plug in the value being approached and expect to get the answer. When we write $$ \lim_{x \to 1}\frac{x^2-1}{x-1}=2 $$ what we mean is that the function $f(x)=\frac{x^2-1}{x-1}$ will get arbitrarily close to $2$ as $x$ approaches $1$. Pick a tiny number—say $0.00001$. As $x$ gets closer and closer to $1$ (but not does ever reach it), $f(x)$ will get within that number of $2$; that is, it will get within the range $(1.99999,2.00001)$. Hopefully this is what you meant when you said 'infinitesimally close'. As you have noticed, sometimes $$ \lim_{x \to a}f(x)=f(a) \, . $$ If this happens, then the function is said to be continuous at $a$. In your example, $$ \lim_{x \to 0}\frac{x}{5}=0 \, . $$ The LHS evaluates to $0$ because as $x$ gets arbitrarily small, so does the value of $x/5$. It just so happens that plugging in $x=0$ also gets you the right answer. However, there are countless examples of functions that aren't continuous. A slightly contrived example is this $$ f(x) = \begin{cases} x+10 &\text{if $x\neq0$} \\ 147 &\text{if $x=0$} \end{cases} $$ The limit as $x$ approaches $0$ of $f(x)$ equals $10$, but $f(0)=147$. The function appears to 'jump' when we get to $x=0$. (Incidentally, this is called a 'jump discontinuity'.) Now, limits to infinity are a different beast altogether. $$ \lim_{x\to\infty}\frac{x^2-1}{x^2+1} = 1 $$ has quite a different meaning to what we have been discussing so far. Here, $\infty$ is just a shorthand for saying that as $x$ gets larger and larger, $$ \frac{x^2-1}{x^2+1} $$ gets closer and closer to $1$. It doesn't make sense for $x$ to get 'arbitrarily close to infinity'—as you have probably been told countless times, infinity is not a number. Consider the sequence $$ \{1,2,3,4,\ldots\} \, . $$ However high you count, you don't ever get 'close' to infinity, but the values in the sequence do increase without bound. That's the sense in which $x$ 'approaches' infinity. And again, $$ \frac{x^2-1}{x^2+1} $$ gets closer and closer to $1$ as $x$ gets bigger and bigger, but does not ever reach it.