Limits by Direct Substitution

Want the temperature of a cooling cup of tea at exactly five minutes? You don't creep up on it second by second — you glance at the thermometer at five minutes and read it off, because the temperature slides along smoothly with no sudden jumps. A huge share of the quantities you meet behave like that, and for them finding a limit is just as easy: read off the value at the point. That is the whole shortcut of this lesson.

For the last few lessons you have been taught to treat limits like unexploded bombs. Approach from the left. Approach from the right. Build a table, watch the values creep in, never touch the point itself. All of that care was honest — a limit really is about the approach, not the arrival — and the limit laws gave the whole business a legal footing.

Now for the enormous relief. For almost every function you will actually meet — polynomials, and fractions whose bottom isn't zero — all that ceremony collapses into a single move: just plug the number in.

\lim_{x \to c} f(x) = f(c).

No table, no graph, no left-and-right sneaking. Substitute x = c, evaluate, done. Functions where this works are called continuous at c: the curve has no break, no jump, no hole there, so where it heads is exactly where it is. The skill of this lesson is not the plugging in — you could do that in year 7. The skill is knowing when you're allowed to, and what to do on the rare occasions the licence is refused.

Why plugging in is legal

This shortcut isn't a new axiom — it's a consequence of the limit laws you already own. Take a polynomial like f(x) = x^2 - 3x + 5 and build its limit from the two atomic facts, \lim_{x \to c} x = c and \lim_{x \to c} k = k. The product law turns \lim x \cdot \lim x into c^2; the constant-multiple law handles the -3x; the sum law stitches the three pieces together:

\lim_{x \to 2} \left(x^2 - 3x + 5\right) = 2^2 - 3(2) + 5 = 3.

Look at what came out: it is exactly the polynomial evaluated at 2. The laws assemble every polynomial the same way, at every point, so every polynomial earns the substitution licence everywhere. And the quotient law extends it to rational functions — a polynomial over a polynomial — at any point where the bottom isn't zero.

Here is the same fact as motion. Drag x toward c = 2 and watch the point ride the curve. Because the curve is unbroken, the ride is uneventful: the point glides straight into (2, 3) with no gap to leap and no cliff to fall off. The value the journey approaches is the value sitting at the destination — that is continuity, drawn.

Try approaching from the left of 2 and from the right. Both journeys end at the same height, f(2) = 3 — which is why, for a continuous function, the careful two-sided check and the lazy plug-in always agree.

If f is continuous at x = c, then \lim_{x \to c} f(x) = f(c) — the limit may be found by direct substitution. (In fact this equation is the definition of continuity at c.)
Polynomials are continuous at every real number. Rational functions \tfrac{p(x)}{q(x)} are continuous wherever the denominator q(c) \neq 0. So are powers, roots (on their domains), \sin x, \cos x and e^x.
Sums, differences, products, quotients (where the bottom is nonzero) and compositions of continuous functions are continuous — anything built from the kit above inherits the licence automatically.

Plug in and read off

Substitution and the graph are two views of one fact. The curve below is f(x) = x^2 - 3x + 5 again. Algebra says: substitute x = 2, get 3. Geometry says: run your finger up from 2 on the axis until you hit the curve, then across to read the height. The dashed guides meet exactly at (2, 3) — both views hand you the same number, because for an unbroken curve the point on the graph and the value the graph approaches are one and the same.

If the curve had a hole punched out at x = 2, the finger-trace would still work — the limit would still be 3 — but substitution would fail, because there would be no f(2) to compute. That's the whole distinction in one picture: the limit reads the neighbourhood, substitution reads the point, and continuity is the promise that they match.

Worked example 1 — a polynomial

Evaluate \lim_{x \to -1} \left(2x^3 - 4x + 7\right).

Step 1 — check the licence. The function 2x^3 - 4x + 7 is a polynomial, and polynomials are continuous at every real number — including -1. Substitution is legal.

Step 2 — substitute. Put x = -1 in, minding the signs:

2(-1)^3 - 4(-1) + 7 = -2 + 4 + 7 = 9.

Step 3 — conclude. \lim_{x \to -1} \left(2x^3 - 4x + 7\right) = 9. Two lines, and the first line is the important one: because the function is continuous there, the value at the point speaks for the limit. Write that justification in an exam — it's the difference between using a theorem and getting lucky.

Worked example 2 — a rational function (check the bottom first)

Evaluate \lim_{x \to 3} \frac{x^2 - 5}{x + 2}.

Step 0 — check the bottom. This is the habit to build. Before touching the numerator, evaluate the denominator at the target: 3 + 2 = 5 \neq 0. The function is a ratio of polynomials with a nonzero bottom at x = 3, so it is continuous there — licence granted.

Step 1 — substitute.

\frac{3^2 - 5}{3 + 2} = \frac{9 - 5}{5} = \frac{4}{5}.

Step 2 — conclude. \lim_{x \to 3} \frac{x^2 - 5}{x + 2} = \tfrac{4}{5}.

The check took three seconds and it is never wasted: when the bottom is nonzero it certifies your answer, and when the bottom is zero it stops you from writing nonsense. Make "evaluate the denominator first" a reflex, the way you check the discriminant before quoting roots.

Worked example 3 — when the licence is refused

Two limits that look just as pluggable, both aiming at x = 2:

\text{(a)}\ \lim_{x \to 2} \frac{x^2 - 4}{x - 2} \qquad\qquad \text{(b)}\ \lim_{x \to 2} \frac{x + 3}{x - 2}.

Step 0 — check the bottom. In both cases the denominator at x = 2 is 2 - 2 = 0. Stop. The substitution licence is refused — the functions aren't even defined at 2, let alone continuous there. But don't walk away yet: the way substitution fails is itself a diagnosis. Check the numerator.

Limit (a): the top is also zero. Substituting gives \tfrac{4 - 4}{2 - 2} = \tfrac{0}{0}. This is the indeterminate form: it does not mean the limit is 0, and it does not mean the limit fails. It means substitution is the wrong tool and the answer is still hiding. (It's hiding in plain sight here: x^2 - 4 = (x-2)(x+2), the offending factor cancels, and the limit turns out to be 4 — the factoring machinery is the subject of indeterminate forms.)

Limit (b): the top is not zero. Substituting gives \tfrac{5}{0} — a nonzero number over zero. This is not indeterminate at all; it is a verdict. Near x = 2 the numerator stays close to 5 while the denominator shrinks toward zero, so the fraction's size explodes: 50, 500, 5000, \ldots The function blows up (a vertical asymptote at x = 2), and the limit genuinely does not exist. No algebra will rescue it — there is nothing to rescue.

Same-looking failure, opposite meanings: \tfrac{0}{0} says "keep working", \tfrac{k}{0} says "case closed, no limit". So the full triage for any limit is:

Substitution returns…	Verdict	What you do
a number L	that number is the limit	done — cite continuity
\tfrac{k}{0} with k \neq 0	the limit does not exist (blow-up)	stop — report a vertical asymptote
\tfrac{0}{0}	no verdict — indeterminate	more algebra: factor and cancel (next lesson)

Substitution is always the first thing to try — not because it always succeeds, but because even its failures tell you exactly what to do next.

Three ways this shortcut gets misused:

Substitution is a licensed shortcut, not the definition. A limit is still, by definition, about the approach: what f(x) closes in on as x nears c, never mind the point itself. f(c) is a different animal — the value at the point. Continuity is precisely the promise that the two agree, and that promise is your licence to plug in. Where the promise fails (a hole, a jump, a zero denominator), the limit may still exist perfectly well — you just can't reach it by substitution.
Getting \tfrac{0}{0} does not mean "no limit". It means substitution can't tell you — the question is above its pay grade. Don't write "undefined" and move on, and never write "= \tfrac{0}{0}" as if it were a value (it isn't a number at all). The limit might be 4, might be 0, might not exist — the form alone can't say, which is exactly why it's called indeterminate. Do more work.
Don't confuse \tfrac{0}{0} with \tfrac{k}{0}. They look like cousins; they are opposites. \tfrac{k}{0} with k \neq 0 is a definite verdict — the function blows up and the limit genuinely fails. \tfrac{0}{0} is the absence of a verdict. Treating a blow-up as "indeterminate" wastes an exam's worth of algebra on a limit that was never there; treating \tfrac{0}{0} as "does not exist" throws away limits that exist beautifully.

It can feel suspicious that the great, careful limit machinery so often reduces to "just plug in". Here's why the world conspires that way. Every function you meet at A-level is assembled from a small kit of parts — powers of x, roots, \sin, \cos, e^x, \ln x — and each part is continuous on its natural domain. Then the assembly operations (add, multiply, divide, compose) all preserve continuity, except at the isolated points where a denominator hits zero. Continuity is hereditary: build from continuous bricks and the whole building is continuous, so substitution works everywhere except at a handful of clearly-flagged danger points.

Genuinely pathological functions do exist — mathematicians build them on purpose. The classic is Dirichlet's monster: the function that equals 1 whenever x is rational and 0 whenever it's irrational. Since every interval, however tiny, contains both kinds of number, the graph is two infinitely-dusty layers and the function is continuous nowhere — no limit exists at any point. But notice what it took to break continuity: a deliberate, hand-crafted definition that no physical process would ever produce. In the calculus you'll actually do — and in the physics and economics it models — something like 95% of the limits you meet fall to substitution, either immediately or after one round of factoring. The care of the last few lessons wasn't wasted: it's what lets you certify the easy cases in one line, and recognise the hard ones on sight.

Watch Sal substitute

Sal Khan runs the same triage: try substitution first, read what comes out, and only reach for heavier tools when the form demands it.