Limit laws — sum, product, quotient, constant multiple: Visual walkthrough

Step 0 — The one word we must define first: "limit"

Before any law, we need to agree what the arrow in $lim_{x \to a} f (x) = L$ actually means. No symbol earns a place until we can point at it.

Term by term, on the picture below:

$a$ — the input we creep toward (on the horizontal axis).
$L$ — the height the outputs are heading to (on the vertical axis).
$ε$ (epsilon) — the half-height of a horizontal band around $L$ . This is your challenge: "get inside this band."
$δ$ (delta) — the half-width of a vertical strip around $a$ . This is my answer: "stay in this strip and you're inside your band."

Step 1 — Set up the two functions we're adding

WHAT. We assume two limits already exist: $lim_{x \to a} f (x) = L lim_{x \to a} g (x) = M .$

$f$ heads to height $L$ ; $g$ heads to height $M$ .
We want the destination of their sum $f + g$ , whose height at each $x$ is $f (x) + g (x)$ .

WHY. The whole trick is that we get to assume $f$ and $g$ are already tame. (If they weren't — see One-sided limits and Mistake B in the parent — the law says nothing.)

PICTURE. Two curves, each sliding into its own band as $x \to a$ . The red curve is the sum $f + g$ ; watch where its height goes.

Step 2 — Name the error, and split it in two

WHAT. The quantity we must shrink is the sum's error — how far $f (x) + g (x)$ is from the target $L + M$ : $total error of the sum (f (x) + g (x)) - (L + M) .$ Here $∣ \cdot ∣$ means distance (always $\geq 0$ ; it forgets the sign). Regroup the inside: $(f (x) + g (x)) - (L + M) = f ’s error (f (x) - L) + g ’s error (g (x) - M) .$

WHY. We can't control the sum's error directly. But we can control $f$ 's error and $g$ 's error separately, because each of those has an existing limit. So we rewrite the big unknown as a combination of two knowns.

PICTURE. The total error is a single vertical gap; below it, that same gap is drawn as two stacked pieces — the $f$ -gap and the $g$ -gap.

Step 3 — The only inequality we need: the triangle inequality

WHAT. For any two numbers $p$ and $q$ : $∣ p + q ∣ \leq ∣ p ∣ + ∣ q ∣.$

$∣ p + q ∣$ — distance you land from $0$ after taking both steps together.
$∣ p ∣ + ∣ q ∣$ — total distance if you took each step and never let them cancel.

Apply it with $p = f (x) - L$ and $q = g (x) - M$ : $(f - L) + (g - M) \leq f ’s error ∣ f - L ∣ + g ’s error ∣ g - M ∣ .$

WHY THIS TOOL, not another? We need an upper bound on the combined error. The danger is cancellation gone wrong — two errors adding up. The triangle inequality is exactly the statement "combined distance is never more than separate distances added": it hands us the worst case for free, so if the worst case is small, we're safe.

PICTURE. Two steps $p$ then $q$ on a number line. The straight-shot distance $∣ p + q ∣$ (red) is never longer than walking $∣ p ∣$ then $∣ q ∣$ (black), and equals it only when both steps point the same way.

Step 4 — Spend your tolerance: give each function half

WHAT. You hand me a target tolerance $ε$ for the sum. I split it in two and spend $\frac{ε}{2}$ on each function.

Because $lim f = L$ exists, I can find a strip half-width $δ_{1}$ forcing $∣ f (x) - L ∣ < \frac{ε}{2}$ .
Because $lim g = M$ exists, I can find a strip half-width $δ_{2}$ forcing $∣ g (x) - M ∣ < \frac{ε}{2}$ .

WHY split as $\frac{ε}{2} + \frac{ε}{2}$ ? Because when I add the two guaranteed errors back, I want them to total exactly $ε$ — no more. Halves are the natural budget: $\frac{ε}{2} + \frac{ε}{2} = ε$ .

PICTURE. The sum's band (height $ε$ ) is split into two half-bands of height $ε /2$ — one budget for $f$ , one for $g$ .

Step 5 — Both strips at once: take the smaller one

WHAT. $δ_{1}$ keeps $f$ in budget; $δ_{2}$ keeps $g$ in budget. To keep both in budget simultaneously, stand inside the narrower strip: $δ = min (δ_{1}, δ_{2}) .$

$min$ — "the smaller of the two." Being inside the smaller strip means you're automatically inside the larger one too.

WHY the minimum? Each guarantee only holds inside its own strip. Their overlap is the narrower strip. Choosing $δ = min$ lands us in the overlap, where both promises are active at once.

PICTURE. Two strips of widths $δ_{1}$ and $δ_{2}$ overlaid; the shaded overlap is the chosen $δ$ -strip.

Step 6 — Snap the pieces together

WHAT. For every $x$ with $0 < ∣ x - a ∣ < δ$ , chain everything:

(f + g) - (L + M) Step 3 \leq < ε /2 ∣ f - L ∣ + < ε /2 ∣ g - M ∣ < \frac{ε}{2} + \frac{ε}{2} = ε .

Reading left to right: split the error (Step 2), bound it by the two separate errors (Step 3), each of which is under $ε /2$ because we're in the $δ$ -strip (Steps 4–5), and the halves rebuild $ε$ .

WHY this finishes it. You named any $ε$ ; I produced a $δ$ that keeps the sum inside your band. That is precisely the definition from Step 0. So: $x \to a lim (f (x) + g (x)) = L + M ■$

PICTURE. The full chain as a flow: total error → two half-errors → each under $ε /2$ → sum under $ε$ .

Step 7 — The edge cases (so no scenario surprises you)

A proof isn't done until every input is covered.

Case A — one limit doesn't exist. Then $L$ or $M$ isn't a number and the whole argument never starts. The law is silent, not false. Classic trap: $f = ⌊ x ⌋$ , $g = - ⌊ x ⌋$ at an integer — each jumps (no limit), yet $f + g \equiv 0$ has limit $0$ . You may not run the sum law backwards. (See One-sided limits.)

Case B — the difference $f - g$ . Same proof with $- g$ in place of $g$ : the constant-multiple idea gives $lim (- g) = - M$ , so $lim (f - g) = L - M$ . No new work.

Case C — adding many functions. Add a third function $h \to N$ : apply the two-function law to $(f + g)$ and $h$ . Since $(f + g) \to L + M$ already, $(f + g) + h \to (L + M) + N$ . Induction extends this to any finite number of pieces — which is exactly why polynomials succumb to plug-in.

Case D — where does $δ$ -smallness fail? Never, for a finite sum. But it can fail for an infinite sum of functions — that's a different theory. Our law is a finite-sum law.

PICTURE. The floor-function counterexample: two functions that jump oppositely, whose sum is a flat red line — proof that "sum has a limit" does not rescue the parts.

The one-picture summary

Everything above, compressed: you name a band of height $ε$ around $L + M$ ; I split it into two half-bands, catch $f$ in one and $g$ in the other using the strips their own limits guarantee, take the narrower strip, and the triangle inequality glues the two captures into one capture inside your full band.

Recall Feynman retelling — explain the whole walkthrough to a friend

Two people are walking toward two chairs. One will sit at height $L$ , the other at height $M$ . I claim their combined height heads to $L + M$ . You're skeptical, so you draw a thin band around $L + M$ and dare me: "make their total land in here." I split your band into two thinner bands — one for each walker — and I already know each walker eventually stays inside their own thin band (that's what "has a limit" means). I wait until both walkers are settled — that means waiting the longer of the two waits, i.e. the narrower strip. Once both are in their half-bands, their total is squeezed into your full band, because the worst two small errors can do when added is still small (the triangle inequality). Since I can beat any band you draw, the total's destination is nailed to $L + M$ . And the fine print: this only works because each walker had a chair to head to — if one wandered forever, all bets are off.

Connections

Epsilon-Delta definition of a limit — Step 0 is this definition; the whole page is one $ε$ – $δ$ game.
One-sided limits — the edge cases (floor function) live here.
Continuity — sums of continuous functions are continuous, straight from this proof.
Squeeze theorem — the sibling tool when the algebra laws can't reach.
Derivative as a limit — the sum rule for derivatives is this law wearing a costume.