What approaching actually means

Consider this function:

Try to evaluate it at $x = 2$. The numerator is $0$, the denominator is $0$, and $0/0$ is undefined. The function has no value at $x = 2$. There is a hole in the graph.

But factor the top: $x^2 - 4 = (x-2)(x+2)$. For every $x$ except 2, the $(x-2)$ cancels and $f(x) = x + 2$. So just to the left and just to the right of the hole, the function is sitting right next to the line $x + 2$, which passes through $(2, 4)$.

The function is undefined at 2, yet everything around 2 is pointing straight at the value 4. That gap, between what happens at a point and what happens near it, is what a limit is for.

Pick the function $(x^2-4)/(x-2)$ in the widget and drag the probe to $x = 2$.

The table evaluates $f$ at points closing in on 2 from both sides: $1.9, 1.99, 1.999$ from the left, and $2.001, 2.01, 2.1$ from the right. Hit zoom to close in tighter.

Read the columns. From the left the outputs march $3.9, 3.99, 3.999, \ldots$. From the right they march $4.1, 4.01, 4.001, \ldots$. Both sides are squeezing in on the same number, 4, even though the function never actually has a value at 2. The table never asks for $f(2)$; it only ever asks what happens nearby.

Here is what that table is measuring.

$\displaystyle\lim_{x \to a} f(x) = L$ means: $f(x)$ gets and stays arbitrarily close to $L$ as $x$ closes in on $a$ from either side, ignoring whatever happens at $a$ itself.

Two phrases carry the weight. From either side: both the left approach and the right approach have to agree. Ignoring whatever happens at $a$: the value $f(a)$ is explicitly not consulted. A hole at $x = a$ does not block a limit. The function does not even have to be defined at $a$.

A limit is a statement about the neighborhood of a point, never about the point itself.

Splitting the approach into two halves gives a sharper tool: the one-sided limits.

• The left limit $\displaystyle\lim_{x\to a^-} f(x)$ watches only $x$ values below $a$.
• The right limit $\displaystyle\lim_{x\to a^+} f(x)$ watches only $x$ values above $a$.

The rule that ties them together:

The two-sided limit exists exactly when both one-sided limits exist and agree. If the two sides head for different numbers, there is no single answer, and the limit does not exist.

Switch the widget to $|x|/x$ and probe $x = 0$.

For any positive $x$, $|x|/x = 1$. For any negative $x$, $|x|/x = -1$. The left column of the table sits flat at $-1$; the right column sits flat at $+1$.

Both one-sided limits exist. They simply disagree. So the two-sided limit at 0 does not exist. The graph has a clean jump, a step down of height 2, and no single value lives in the gap. A tempting wrong move is to average the two sides and call the limit 0. Resist it: the function never goes near 0; it is only ever exactly $-1$ or exactly $+1$.

Cycle the widget through its remaining functions. There are exactly three ways a two-sided limit can fail.

1. Jump. The one-sided limits disagree, like $|x|/x$ at 0.
2. Blow-up. The function runs off to $\pm\infty$, like $1/x^2$ at 0. Note carefully: “the limit is $+\infty$” is a named failure, not a value. The limit still does not exist.
3. Oscillation. Try $\sin(1/x)$ near 0. As $x$ shrinks, $1/x$ races off to infinity, so $\sin(1/x)$ swings between $-1$ and $+1$ faster and faster, forever. Zoom in and the wiggles get denser, never calmer. No jump, no asymptote, and still no limit, because the output never settles on anything.

That third one is the subtle case. The input $x$ heads obediently to 0, but the output refuses to.

Most of the time none of this drama happens. For a well-behaved function, the limit at $a$ is simply $f(a)$.

That well-behaved property has a name. A function $f$ is continuous at $a$ when:

The limit exists, the function is defined there, and the two match. Geometrically: no hole, no jump, no asymptote. You could draw the graph through $a$ without lifting your pen. The three failure modes from the last step are exactly the three ways this equation can break.

Here is the payoff, and it is large.

Every parent function from the first lesson is continuous everywhere on its natural domain. Polynomials, $\sin$, $\cos$, $e^x$, $\sqrt{x}$, $\ln x$, all of them. And continuity survives addition, multiplication, and division (as long as you do not divide by zero).

So for any function built by combining catalog functions, evaluating a limit is just plugging in. Take:

Numerator and denominator are sums of catalog functions, and the denominator is not zero at $x = 1$. So the whole thing is continuous at 1, and the limit is just the value there: $\dfrac{\ln 1 + e^1}{1 + 3} = \dfrac{0 + e}{4} = \dfrac{e}{4}$.

No factoring, no tables, no tricks. Reach for tables and zooming only when plugging in gives you $0/0$ or worse.

A transformer’s forward pass is one long composition of catalog functions stacked into a computation graph. Because every piece is continuous, the whole graph is continuous, and that is not a technicality. Backpropagation, the algorithm that trains the network, only works because each piece varies smoothly enough to have a slope.

You now have the limit. The last lesson of this module turns it on a curve and watches a slope appear.