Functions, limits, and continuity: Visual walkthrough

This is a visual walkthrough of the parent idea in Functions, limits, and continuity. We rebuild the central result — the limit of $\frac{x ^{2} - 1}{x - 1}$ as $x \to 1$ equals $2$ , even though the machine gives $0/0$ there — from absolute zero, one picture per step. By the end you will see why a hole in the graph does not stop us from knowing where the curve is heading.

Step 1 — What "function" means as a picture

WHAT. We start with the machine $f (x) = \frac{x ^{2} - 1}{x - 1}$ . Feed it a number $x$ , it spits out a number.

WHY. Before we can ask "where is the output heading (i.e. what is its limit)?" we must be able to see the output. So we plot: horizontal axis = the input $x$ , vertical axis = the output $f (x)$ . Every input becomes a point on a curve.

PICTURE. Look at the cyan curve in the figure. The input $x$ runs left–right; the output $f (x)$ runs up–down. For each $x$ we drop a dot at height $f (x)$ . Notice there is nothing special-looking yet — it just traces a straight slanted line. That surprise is the whole story.

Figure — Functions, limits, and continuity

Step 2 — Where the machine chokes: $0/0$

WHAT. Put $x = 1$ straight into the machine.

WHY. We must find exactly where the function is undefined, because that is the only interesting spot — everywhere else you can just plug in and read the answer.

$f (1) = \frac{1 ^{2} - 1 = 0}{= 0 1 - 1} = \frac{0}{0}$

Term by term: the top $1^{2} - 1$ becomes $0$ ; the bottom $1 - 1$ becomes $0$ . So we are asked to compute $0 \div 0$ .

WHY this is not "the limit fails". Division wants to answer "how many bottoms fit in the top?" With $0/0$ that question has no single answer — any number times $0$ gives $0$ , so nothing is pinned down. This is called an indeterminate form. It is a "please simplify me" sign, not a verdict.

PICTURE. The amber dot marks the input $x = 1$ on the axis, and the amber hollow circle on the curve shows: right there, at height where the curve would be, the machine has no value. The curve has a hole.

Step 3 — Approach from BOTH sides with numbers

WHAT. Instead of landing on $x = 1$ , we creep toward it — from the left ( $x < 1$ ) and from the right ( $x > 1$ ) — and watch the output.

WHY. The limit only cares about the journey toward the point, never the destination at the point. So we sneak up to $1$ without ever touching it, from each side, and see if both sides aim at the same height.

Approaching from the LEFT ( $x < 1$ , climbing up toward $1$ ):

$x$ (from left)	$f (x)$
$0.9$	$1.9$
$0.99$	$1.99$
$0.999$	$1.999$

Approaching from the RIGHT ( $x > 1$ , settling down toward $1$ ):

$x$ (from right)	$f (x)$
$1.1$	$2.1$
$1.01$	$2.01$
$1.001$	$2.001$

Both tables squeeze toward $2$ . The left side climbs up to $2$ ; the right side settles down to $2$ .

PICTURE. Two amber arrows walk along the curve toward the hole — one from the left, one from the right. Their heights both funnel to the dashed line at $y = 2$ . The hole sits exactly where they meet, but neither arrow needs to step on it.

Step 4 — WHY both sides agree: factor the top

WHAT. We simplify the algebra to explain the numbers, not just observe them. Factor the numerator.

WHY. The difference of squares lets us pull the troublesome $x - 1$ out of the top. We use this tool because the bottom is exactly $x - 1$ — if we can copy that factor in the top, they cancel.

$x^{2} - 1 = x^{2} - 1^{2} = (x - 1) (x + 1)$

Here $a = x$ , $b = 1$ , so $a - b = x - 1$ and $a + b = x + 1$ . The factor $(x - 1)$ is the twin of our denominator.

PICTURE. The figure shows the top curve $x^{2} - 1$ splitting into two straight-line factors $(x - 1)$ (cyan) and $(x + 1)$ (white); both cross zero, but only $x - 1$ shares its zero with the bottom.

Step 5 — The legal cancellation

WHAT. Cancel the shared $(x - 1)$ .

$\frac{x ^{2} - 1}{x - 1} = \frac{( x - 1 ) ( x + 1 )}{x - 1} = x + 1 (valid for x \neq = 1)$

WHY it is allowed. Dividing top and bottom by $(x - 1)$ is only legal when $x - 1 \neq = 0$ , i.e. when $x \neq = 1$ . And that is exactly the region a limit lives in — the strict inequality $0 < ∣ x - 1∣$ in the ( $ε, δ$ ) definition excludes $x = 1$ . So on the whole journey (everywhere except the forbidden point), the ugly fraction is identical to the simple line $x + 1$ .

PICTURE. The same slanted line $x + 1$ is drawn, with the single hole at $(1, 2)$ redrawn on top. The message: the complicated fraction and the tidy line are the same curve apart from one punched-out point.

Step 6 — Read off the answer

WHAT. Now $x + 1$ is a friendly straight line with no hole in its formula, so its limit at $x = 1$ is just its value there.

$lim_{x \to 1} (x + 1) = 1 + 1 = 2$

Term by term: as $x \to 1$ , the piece $x \to 1$ and the piece $+ 1$ stays $+ 1$ , so the sum heads to $2$ .

WHY plugging in is now legal. $x + 1$ is continuous (see Continuity of composite functions): no hole, no jump, no cliff. For continuous functions the limit equals the value — that is what continuity means. So and only so may we substitute.

PICTURE. The final read-off: a horizontal dashed line at $y = 2$ meets the curve at the (open) target point. The two approaching arrows land there. The limit is $2$ .

Step 7 — Edge case: what if the two sides disagreed?

WHAT. We must also show the case where a limit does not exist, so you can tell them apart.

WHY. A picture of success is only convincing next to a picture of failure. Take the step function $s (x) = {03 x < 1 x > 1$

WHY it fails. From the left the height sits at $0$ ; from the right it jumps to $3$ . Left-limit $= 0$ , right-limit $= 3$ , and $0 \neq = 3$ — the two-sided limit does not exist. This is a jump discontinuity, a genuine break, unlike our removable hole.

PICTURE. One arrow aims at height $0$ , the other at height $3$ ; they never meet, so there is no single number to call the limit. Contrast this with Step 3, where both aimed at the same $2$ .

Step 8 — What "patching the hole" actually means

WHAT. We now repair $f$ so there is no hole. To redefine a function at a point means: keep the same rule everywhere except at that one input, and there declare a new output by hand. Formally, we build a new function $\tilde{f} (x) = ⎩ ⎨ ⎧ \frac{x ^{2} - 1}{x - 1} 2 x \neq = 1 x = 1$

Term by term: the top line is the old rule, used for every $x \neq = 1$ ; the bottom line is our hand-placed value $\tilde{f} (1) = 2$ , filling the spot the old machine left blank.

WHY the value $2$ and not any other number. We choose $2$ because that is the limit (Steps 3–6). With that choice all three continuity conditions hold: $\tilde{f} (1)$ exists ( $= 2$ ), $lim_{x \to 1} \tilde{f} (x)$ exists ( $= 2$ ), and they are equal. So $\tilde{f}$ is continuous at $1$ — the pen no longer lifts.

PICTURE. The open circle from Step 5 is filled in solid at $(1, 2)$ : same line, hole plugged.

The one-picture summary

Everything in one frame: the ugly fraction $\frac{x ^{2} - 1}{x - 1}$ is the straight line $x + 1$ with a single point punched out at $(1, 2)$ . Approach from both sides, both aim at $y = 2$ , and since we never step on the hole, the limit is $2$ . Only when we redefine $\tilde{f} (1) := 2$ (fill the hole with the limit value) does the function become continuous — limit equals value.

Recall Feynman: the whole walkthrough in plain words

You want to know where a curve is heading at one spot, but the machine breaks there and yells " $0/0$ ". Don't panic — that yell just means "simplify me". You notice the top $x^{2} - 1$ secretly contains a copy of the bottom $x - 1$ , because $x^{2} - 1 = (x - 1) (x + 1)$ . Everywhere except the broken spot, you cancel that copy and the scary fraction turns into the plain line $x + 1$ . Since a limit is only about the journey toward the point and never the point itself, the broken spot never gets visited — so the fraction and the line have the exact same limit. That limit is $1 + 1 = 2$ . To check you're right, you tiptoe in from both sides with numbers ( $0.9, 0.99, \dots$ and $1.1, 1.01, \dots$ ) and both sides squeeze toward $2$ . A hole like this is removable: patch it by declaring the value at that one point to be $2$ — that is, build a new function identical everywhere except set its value at $1$ equal to $2$ — and the pen never lifts again. Compare a jump (like a step function): there the two sides aim at different heights, they never agree, and no patch can save it.

Connections

Derivatives and gradients — the derivative is itself a limit of a fraction that hits $0/0$ ; this exact cancellation trick is how you compute slopes by hand.
L'Hôpital's Rule — a power tool for the same $0/0$ problem when factoring is too hard.
Continuity of composite functions — why $x + 1$ lets us "just plug in".
Activation functions (ReLU, sigmoid) — ReLU has a kink, not a hole or jump; continuous but not smooth.
Gradient Descent — needs these limits to behave so slopes exist.