WaywardHouse

27  Limits and continuity

What a function approaches, not what it equals

Your car’s speedometer reads 80 km/h at this exact instant. But speed is distance divided by time — and at a single instant, no time has passed, no distance covered. So what is the speedometer actually measuring?

The answer is a limit. It’s the value that the ratio distance/time approaches as the time interval shrinks toward zero — not the ratio at zero, which is 0/0 and meaningless, but the value that ratio gets closer and closer to.

You’ve already seen this pattern in another form. In the last chapter you watched sequences converge to numbers they never quite reached. A limit is that same idea applied to functions.

27.1 The idea before the notation

Three situations will make this concrete. Notice what they have in common.

The speedometer. A car’s position at time \(t\) is some function \(s(t)\). To estimate speed, you record position over a short interval: speed \(\approx (s(t + \Delta t) - s(t)) / \Delta t\). The shorter the interval, the better the estimate. The true instantaneous speed is what this ratio approaches as \(\Delta t\) shrinks toward zero. The ratio itself is never evaluated at \(\Delta t = 0\) — that would be \(0/0\) — but it heads steadily toward a definite value.

The temperature probe. A weather sensor is mounted on a boundary between two air masses — one at 18°C and one at 12°C. On the warm side, readings approach 18°C. On the cold side, readings approach 12°C. The sensor itself sits exactly at the boundary and might read either, neither, or something else entirely depending on how it was calibrated. The function has a well-defined approach from each side. Whether the value at the boundary matches either approach is a separate question.

Sequences converging to a number. In Vol 4 ch 4 you watched the sequence \(1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots\) converge to zero. The terms never equal zero, but they get arbitrarily close. Functions can do exactly the same thing: a function can get arbitrarily close to a value \(L\) as \(x\) approaches some point \(a\), without ever equalling \(L\) at \(x = a\).

All three situations share the same structure: a value is approached without necessarily being reached. That is the concept of a limit.

27.2 What the notation is saying

The symbol

\[\lim_{x \to a} f(x) = L\]

is read: the limit of \(f(x)\) as \(x\) approaches \(a\) equals \(L\).

Every piece matters:

  • \(x \to a\) means \(x\) gets closer and closer to \(a\), from either side, but does not equal \(a\).
  • \(f(x)\) is the output of the function at each nearby \(x\).
  • \(L\) is the value that \(f(x)\) approaches.

The limit says nothing about \(f(a)\) itself. The function might be defined at \(a\), undefined at \(a\), or defined but with the wrong value. The limit only asks about the approach.

27.2.1 One-sided limits

Sometimes the approach from the left and the approach from the right give different values. Two separate notations handle this.

\[\lim_{x \to a^-} f(x)\]

is the left-hand limit — the value \(f(x)\) approaches as \(x\) comes in from values less than \(a\). Read it: the limit from the left.

\[\lim_{x \to a^+} f(x)\]

is the right-hand limit — the value \(f(x)\) approaches as \(x\) comes in from values greater than \(a\). Read it: the limit from the right.

The two-sided limit exists if and only if both one-sided limits exist and are equal. If the left-hand limit is \(L_1\) and the right-hand limit is \(L_2\), then:

\[\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L\]

If \(L_1 \neq L_2\), the two-sided limit does not exist.

27.2.2 The limit versus the value — three cases

There is no rule that says \(\lim_{x \to a} f(x)\) must equal \(f(a)\). In fact, that equality is the definition of continuity, and it can fail in three genuinely distinct ways.

Case 1 — Continuous point. The function is defined at \(a\), the limit exists, and they agree:

\[\lim_{x \to a} f(x) = f(a)\]

Example: \(f(x) = x^2\) at \(a = 3\). The limit is 9, the value is 9.

Case 2 — Removable discontinuity (value mismatch). The limit exists, \(f(a)\) is defined, but they are unequal. The function has a value at \(a\) — it is simply the wrong value relative to what the surrounding behaviour predicts.

Example: define \(g(x) = x + 2\) for \(x \neq 2\), but \(g(2) = 7\). The limit as \(x \to 2\) is 4. The function value at \(x = 2\) is 7. There is a mismatch. The discontinuity can be removed by redefining \(g(2) = 4\).

Case 3 — Hole (value absent). The limit exists, but \(f(a)\) is simply not defined. There is no value to mismatch — the function has a gap at \(x = a\).

Example: \(h(x) = \dfrac{x^2 - 4}{x - 2}\). At \(x = 2\) the denominator is zero, so \(h(2)\) is undefined. But for every \(x \neq 2\):

\[h(x) = \frac{(x-2)(x+2)}{x-2} = x + 2\]

The limit is 4. The function has a hole at \((2, 4)\) — no value is there at all. Patching the hole by defining \(h(2) = 4\) would make \(h\) continuous.

Cases 2 and 3 are genuinely different situations: in Case 2 you have a value but it conflicts with the limit; in Case 3 you simply have nothing. Both are called removable discontinuities because the fix — redefining or defining \(f(a) = L\) — is the same in both cases.

27.3 The method — evaluating limits

Four techniques cover the vast majority of problems you will encounter.

27.3.1 Technique 1: Direct substitution

If \(f\) is a polynomial, a rational function with a non-zero denominator at \(a\), a trigonometric function, or any other function that is continuous at \(a\) — we’ll define exactly what this means shortly, but the idea is that the function has no break at that point — then:

\[\lim_{x \to a} f(x) = f(a)\]

Just substitute. This always works when no algebraic obstruction exists.

When to use it: Try this first. If the substitution gives a real number, you’re done.

Worked example: Find \(\lim_{x \to 3} (x^2 + 2x - 1)\).

Substitute \(x = 3\) directly:

\[\lim_{x \to 3} (x^2 + 2x - 1) = 3^2 + 2(3) - 1 = 9 + 6 - 1 = 14\]

The limit is 14. The function is a polynomial — continuous everywhere — so direct substitution is valid at every real number.

When it fails: If substitution gives \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), or another indeterminate form, you need one of the other techniques.

27.3.2 Technique 2: Factor and cancel

The \(\frac{0}{0}\) form is a signal: both numerator and denominator share a common factor that vanishes at \(x = a\). Factor both, cancel the common factor, and then substitute.

When to use it: When direct substitution gives \(\frac{0}{0}\) and the expression is a ratio of polynomials.

Worked example: Find \(\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}\).

Step 1. Attempt direct substitution: \(\dfrac{2^2 - 4}{2 - 2} = \dfrac{0}{0}\). Indeterminate — proceed to step 2.

Step 2. Factor the numerator:

\[\frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2}\]

Step 3. Cancel the common factor \((x - 2)\). This is valid because we are evaluating a limit as \(x \to 2\), meaning \(x \neq 2\), so \((x-2) \neq 0\):

\[\frac{(x-2)(x+2)}{x-2} = x + 2 \quad (x \neq 2)\]

Step 4. Now substitute:

\[\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2}(x + 2) = 2 + 2 = 4\]

The cancellation is legal only because we never set \(x = 2\) — the limit asks what happens as \(x\) approaches 2, not at \(x = 2\).

27.3.3 Technique 3: Rationalise

When a square root appears and direct substitution gives \(\frac{0}{0}\), the standard move is to multiply numerator and denominator by the conjugate of the expression containing the square root. The conjugate of \((\sqrt{u} - v)\) is \((\sqrt{u} + v)\). Multiplying removes the radical from one factor using the difference-of-squares identity.

When to use it: When the \(\frac{0}{0}\) form arises from a square root.

Worked example: Find \(\lim_{x \to 0} \dfrac{\sqrt{x+4} - 2}{x}\).

Step 1. Direct substitution: \(\dfrac{\sqrt{0+4} - 2}{0} = \dfrac{2 - 2}{0} = \dfrac{0}{0}\). Indeterminate — proceed.

Step 2. Multiply numerator and denominator by the conjugate \((\sqrt{x+4} + 2)\):

\[\frac{\sqrt{x+4} - 2}{x} \cdot \frac{\sqrt{x+4} + 2}{\sqrt{x+4} + 2}\]

Step 3. Expand the numerator using \((a-b)(a+b) = a^2 - b^2\):

\[\frac{(\sqrt{x+4})^2 - 2^2}{x(\sqrt{x+4} + 2)} = \frac{(x+4) - 4}{x(\sqrt{x+4} + 2)} = \frac{x}{x(\sqrt{x+4} + 2)}\]

Step 4. Cancel the common factor \(x\) (valid since \(x \to 0\) means \(x \neq 0\)):

\[\frac{x}{x(\sqrt{x+4} + 2)} = \frac{1}{\sqrt{x+4} + 2}\]

Step 5. Now substitute \(x = 0\):

\[\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x} = \frac{1}{\sqrt{0+4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}\]

The limit is \(\frac{1}{4}\).

27.3.4 Technique 4: One-sided limits from graphs

When you have a graph (or can sketch one), you can read limits directly. Look at the curve’s approach as \(x\) moves toward the target from each side. Do not look at the dot at the target point — look at where the curve is heading.

When to use it: Piecewise functions, graphs given without formulas, or any function with a potential jump.

Worked example: Let

\[f(x) = \begin{cases} x + 1 & x < 3 \\ 5 & x = 3 \\ 2x - 4 & x > 3 \end{cases}\]

Find \(\lim_{x \to 3^-} f(x)\), \(\lim_{x \to 3^+} f(x)\), and \(\lim_{x \to 3} f(x)\).

Left-hand limit. For \(x < 3\), the rule is \(f(x) = x + 1\). As \(x \to 3^-\):

\[\lim_{x \to 3^-} f(x) = 3 + 1 = 4\]

Right-hand limit. For \(x > 3\), the rule is \(f(x) = 2x - 4\). As \(x \to 3^+\):

\[\lim_{x \to 3^+} f(x) = 2(3) - 4 = 2\]

Two-sided limit. The left and right limits are \(4 \neq 2\), so:

\[\lim_{x \to 3} f(x) \text{ does not exist}\]

Note that \(f(3) = 5\) is defined — but the two-sided limit still fails to exist because the one-sided limits disagree. The value at the point is irrelevant to whether the limit exists.

27.4 Continuity

With limits in hand, the definition of continuity is short.

Definition. A function \(f\) is continuous at \(a\) if all three conditions hold:

  1. \(f(a)\) is defined.
  2. \(\lim_{x \to a} f(x)\) exists.
  3. \(\lim_{x \to a} f(x) = f(a)\).

If any condition fails, \(f\) is discontinuous at \(a\).

Informally: a function is continuous at a point if you can draw through that point without lifting your pen. The limit and the value agree.

27.4.1 The three types of discontinuity

Removable discontinuity. Condition 1 or 3 fails, but condition 2 holds. The limit exists, but either the function is not defined there, or the function is defined but gives a different value. The graph has a single isolated hole. The discontinuity can be removed by redefining \(f(a) = L\).

Example: \(g(x) = \dfrac{x^2 - 4}{x - 2}\) is not defined at \(x = 2\), but \(\lim_{x \to 2} g(x) = 4\). Removable.

Jump discontinuity. The left-hand and right-hand limits both exist, but they are not equal. The function “jumps” from one value to another. No amount of redefining a single point can fix this — the gap is structural.

Example: The piecewise function from Technique 4 above. At \(x = 3\), the left limit is 4 and the right limit is 2.

Infinite discontinuity. One or both one-sided limits are \(\pm\infty\). The graph has a vertical asymptote at \(x = a\).

Example: \(h(x) = \dfrac{1}{x}\) at \(x = 0\). As \(x \to 0^+\), \(h(x) \to +\infty\). As \(x \to 0^-\), \(h(x) \to -\infty\).

27.5 The epsilon-delta definition

The techniques above let you compute limits. The epsilon-delta definition lets you prove they are correct. These are different activities, and most students — even in university calculus — use the techniques far more often than the proof machinery. But understanding the definition is what separates a practitioner from a mathematician.

Return to the speedometer. Suppose you want to estimate speed to within \(\varepsilon = 0.01\) km/h. You need to take a time interval short enough — call it \(\delta\) seconds — that the average speed over that interval lands within \(0.01\) km/h of the true instantaneous speed. Different tolerance \(\varepsilon\), different required \(\delta\). The claim “the limit is \(L\)” means: for every tolerance you name, I can find a window small enough to guarantee that accuracy.

Epsilon-delta definition of a limit.

\[\lim_{x \to a} f(x) = L\]

means: for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that

\[0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon\]

In plain language: however small a target band you set around \(L\) (the \(\varepsilon\)-band), I can find a window around \(a\) (the \(\delta\)-window) such that every \(x\) inside the window — but not equal to \(a\) itself — sends \(f(x)\) inside the target band.

One worked example. Prove that \(\lim_{x \to 2} 3x = 6\).

We need to show that for any \(\varepsilon > 0\) we can produce a \(\delta > 0\) satisfying the definition.

We want \(|f(x) - L| < \varepsilon\), i.e., \(|3x - 6| < \varepsilon\).

Factor: \(|3x - 6| = 3|x - 2|\).

So the condition \(|3x - 6| < \varepsilon\) becomes \(3|x - 2| < \varepsilon\), i.e., \(|x - 2| < \varepsilon/3\).

This tells us to choose \(\delta = \varepsilon/3\).

Verification. Suppose \(0 < |x - 2| < \delta = \varepsilon/3\). Then:

\[|3x - 6| = 3|x - 2| < 3 \cdot \frac{\varepsilon}{3} = \varepsilon\]

The condition is satisfied for every \(\varepsilon > 0\). The limit is 6. \(\square\)

Why this works — the logical structure.

The phrase “for every \(\varepsilon > 0\)” covers every possible tolerance, however tiny. The phrase “there exists \(\delta > 0\)” says you can always respond to any challenge.

Together they convert the informal idea of “approaches without reaching” — which involves an infinite process — into a finite, checkable algebraic statement. You never have to evaluate an infinite process; you just have to exhibit the \(\delta\) as a function of \(\varepsilon\) and verify the implication.

This is the logical structure of most rigorous analysis. The techniques in the previous section are shortcuts valid because the functions involved are “nice enough” that the \(\varepsilon\)-\(\delta\) machinery can be applied once and the results packaged into rules.

27.6 Worked examples

27.6.1 Example 1 — Direct substitution (polynomial)

Problem. Find \(\lim_{x \to 3} (x^2 + 2x - 1)\).

Context. In structural engineering, polynomial functions describe deflection curves along a beam. Evaluating a deflection at a specific point is direct substitution.

Solution.

The function \(f(x) = x^2 + 2x - 1\) is a polynomial. Polynomials are continuous everywhere. Direct substitution is valid at \(x = 3\):

\[\lim_{x \to 3}(x^2 + 2x - 1) = 3^2 + 2(3) - 1 = 9 + 6 - 1 = \boxed{14}\]

Technique used: Direct substitution. Applied whenever the function is continuous at the target point — no denominator that vanishes, no radical causing trouble.

27.6.2 Example 2 — Factoring (removable discontinuity)

Problem. Find \(\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}\).

Context. In software engineering, a function representing computational cost might be undefined at a specific input size but still have a meaningful limiting value. Analysing the limit is the correct tool.

Solution.

Step 1. Direct substitution gives \(\frac{0}{0}\) — indeterminate.

Step 2. Factor:

\[\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2}\]

Step 3. Cancel \((x-2)\), valid since \(x \neq 2\) in a limit:

\[= x + 2 \quad (x \neq 2)\]

Step 4. Substitute:

\[\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2}(x + 2) = 2 + 2 = \boxed{4}\]

The function has a hole at \((2, 4)\) — a removable discontinuity. The limit exists and equals 4 even though the function is undefined at \(x = 2\).

27.6.3 Example 3 — Rationalisation (resolving an indeterminate form)

Problem. Find \(\lim_{x \to 0} \dfrac{\sqrt{x+4} - 2}{x}\).

Context. In chemistry, reaction rate expressions often involve differences of concentrations under a square root. Finding the limiting rate at a reference condition uses exactly this technique.

Solution.

Step 1. Direct substitution: \(\frac{\sqrt{4} - 2}{0} = \frac{0}{0}\). Indeterminate.

Step 2. Multiply by the conjugate \(\frac{\sqrt{x+4} + 2}{\sqrt{x+4} + 2}\):

\[\frac{\sqrt{x+4} - 2}{x} \cdot \frac{\sqrt{x+4} + 2}{\sqrt{x+4} + 2} = \frac{(x+4) - 4}{x(\sqrt{x+4} + 2)} = \frac{x}{x(\sqrt{x+4} + 2)}\]

Step 3. Cancel \(x\) (\(x \neq 0\) in the limit):

\[= \frac{1}{\sqrt{x+4} + 2}\]

Step 4. Substitute \(x = 0\):

\[\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x} = \frac{1}{\sqrt{4} + 2} = \frac{1}{4} = \boxed{0.25}\]

27.6.4 Example 4 — One-sided limits (regime change)

Problem. A continuous-time financial model uses the interest rate function

\[r(t) = \begin{cases} 0.04 & t < 5 \\ 0.04 & t = 5 \\ 0.065 & t > 5 \end{cases}\]

where \(t\) is years from the start of a contract. Does \(\lim_{t \to 5} r(t)\) exist?

Context. In finance, many contracts include rate resets at specified times. Whether the limit exists at the reset point determines whether the model is mathematically continuous or whether a jump discontinuity must be handled separately.

Solution.

Left-hand limit: As \(t \to 5^-\), the rule gives \(r(t) = 0.04\):

\[\lim_{t \to 5^-} r(t) = 0.04\]

Right-hand limit: As \(t \to 5^+\), the rule gives \(r(t) = 0.065\):

\[\lim_{t \to 5^+} r(t) = 0.065\]

Two-sided limit: The one-sided limits are unequal (\(0.04 \neq 0.065\)), so:

\[\lim_{t \to 5} r(t) \text{ does not exist}\]

This is a jump discontinuity. Note that \(r(5) = 0.04\) is defined — the value exists at the point — but the two-sided limit does not. The model has a structural break at \(t = 5\) that must be accounted for, not patched over.

27.7 Where this goes

Toward differential calculus (Chapter 2). The derivative of \(f\) at a point \(a\) is defined as:

\[f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\]

This is precisely a limit of the \(\frac{0}{0}\) form — the same kind you evaluated in Examples 2 and 3 above. The technique of factoring, cancelling, and substituting that you practised here is exactly the algebra you will apply to evaluate derivatives in the next chapter. Limits are not preliminary material to get past — they are the mechanism the derivative is made of.

Back to sequences and series (Vol 4 Ch 4). The convergence you studied there was the same idea in a different setting. When you showed that \(\sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^k = 2\), you were evaluating a limit of partial sums: \(\lim_{n \to \infty} S_n = 2\). The same idea — for any tolerance, there’s a point beyond which the partial sums stay within that tolerance — reappears in the formal definition of series convergence. The concepts are one family, not two separate topics.

Where limits appear in practice

  • Speedometers and velocity. Instantaneous velocity is \(\lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t}\) — the starting point for all of mechanics.

  • Continuously compounded interest. \(\lim_{n \to \infty}\left(1 + \frac{r}{n}\right)^n = e^r\). Every continuous-time financial model uses this limit.

  • Numerical root-finding. The bisection method generates a sequence of intervals that converge to a root. Convergence is a limit: each iteration halves the error, and the sequence of midpoints limits to the root.

  • Signal processing. A low-pass filter’s attenuation as frequency \(\to \infty\) is a limit question. Whether the filter “works” depends on whether that limit is zero.

27.8 Exercises

Exercise 1. A structural engineer models the deflection of a beam using \(d(x) = 2x^3 - 3x^2 + x + 1\), where \(x\) is position in metres.

Find \(\lim_{x \to 2} d(x)\).

Exercise 2. A computing platform charges users based on data processed. The cost function has the form

\[C(x) = \frac{x^2 - 9}{x - 3}\]

where \(x\) is gigabytes processed. The pricing model is undefined at \(x = 3\) GB due to a contract edge case. Find \(\lim_{x \to 3} C(x)\).

Exercise 3. In biochemistry, a reaction rate expression takes the form

\[R(c) = \frac{\sqrt{c + 9} - 3}{c}\]

where \(c\) is the change in concentration from a reference value. Find \(\lim_{c \to 0} R(c)\).

Exercise 4. A piecewise model for a climate variable switches regime at time \(t = 10\) years:

\[T(t) = \begin{cases} 2t + 1 & t < 10 \\ 30 & t = 10 \\ 3t - 8 & t > 10 \end{cases}\]

  1. Find \(\lim_{t \to 10^-} T(t)\).
  2. Find \(\lim_{t \to 10^+} T(t)\).
  3. Does \(\lim_{t \to 10} T(t)\) exist? Explain.
  4. Is \(T\) continuous at \(t = 10\)?

Exercise 5. Use the three conditions to determine whether

\[f(x) = \begin{cases} x^2 - 1 & x \neq 2 \\ 3 & x = 2 \end{cases}\]

is continuous at \(x = 2\).

Exercise 6 (epsilon-delta). Use the \(\varepsilon\)-\(\delta\) definition to prove that \(\lim_{x \to 1} (5x - 2) = 3\).