Limits & Continuity — Maths for Thermodynamics & Kinetics

Getting arbitrarily close

The idea that makes calculus rigorous

A limit is the value a function heads toward as its input approaches some point — regardless of whether the function is actually defined there. We write this as:

x \to a lim f (x) = L

If the left-hand and right-hand approaches agree, the limit exists. A function is continuous at $a$ if that limit equals $f (a)$ — no holes, jumps or breaks. Continuity is what allows us to differentiate and integrate.

In chemistry, limiting behaviour appears constantly. As $t \to \infty$ , the concentration of a reactant in a first-order reaction approaches zero. As the volume of a gas approaches infinity (infinite dilution), the pressure approaches zero. As $T \to 0$ , the heat capacity of a crystalline solid approaches zero — and the third law of thermodynamics is built on that limiting statement.

The difference quotient $Δ f /Δ x$ gives the average slope of a chord. Sending $Δ x \to 0$ turns that average into an instantaneous value — the derivative. So every derivative is, at heart, a limit:

f^{'} (x) = Δ x \to 0 lim \frac{f ( x + Δ x ) - f ( x )}{Δ x}

Existence condition

Both one-sided limits must agree:

lim_{x \to a^{-}} f = lim_{x \to a^{+}} f = L

Continuity at a

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

— limit equals the actual value.

Limit as t → ∞

lim_{t \to \infty} [A]_{0} e^{- k t} = 0

— first-order reaction goes to completion.

Difference quotient

The derivative is the limit of

Δ [A] /Δ t

Δ t \to 0

Why chemists care

Both the derivative (a limit of a slope) and the integral (a limit of a sum) are built from limits. And limiting behaviour — “as

t \to \infty

”, “at infinite dilution”, “as

T \to 0

” — describes where chemical systems end up. Open the limit explorer.

There are two distinct types of discontinuity worth recognising. A removable discontinuity (hole) occurs when the limit exists but the function is either undefined or has the wrong value at that point — for example $(x^{2} - 1) / (x - 1)$ at $x = 1$ . The hole can be “filled in” by redefining the function at one point. A jump discontinuity occurs when the left-hand and right-hand limits both exist but differ — as in the step-function titration model. Jump discontinuities cannot be removed by redefining a single value. The distinction matters in kinetics: a sudden change in rate constant at a phase boundary is a jump, whereas a removable singularity in a rate expression often signals a factorable algebraic form.

In consecutive reactions $A k_{1} B k_{2} C$ the concentration of the intermediate is $[B] (t) = \frac{k _{1} [ A ] _{0}}{k _{2} - k _{1}} (e^{- k_{1} t} - e^{- k_{2} t})$ . When $k_{2} \to k_{1}$ this appears to become $0/0$ , but L'Hôpital's rule resolves it to $[B] (t) = k_{1} [A]_{0} t e^{- k_{1} t}$ — finite and physically meaningful at all times. As $t \to \infty$ , even this special-case expression goes to zero: the intermediate is eventually consumed completely.

Removable discontinuity

Limit exists; function value missing or wrong. Repair: redefine at the point.

Jump discontinuity

One-sided limits exist but differ. Cannot be removed. Example: step-function pH model at equivalence point.

L'Hôpital condition

Only apply if both

f (a) = 0

and

g (a) = 0

(or both

\pm \infty

). One check per application.

t \to \infty

in kinetics

Any

e^{- k t}

with

k > 0

vanishes, driving first-order concentrations to equilibrium or zero.

Common pitfalls

Substituting into an indeterminate form. Writing $lim_{x \to 1} \frac{x ^{2} - 1}{x - 1} = \frac{0}{0}$ and stopping there is wrong — the limit is 2, not undefined. Always factor, simplify, or apply L'Hôpital.
Applying L'Hôpital without checking the condition. The rule requires a genuine $0/0$ or $\infty/\infty$ form. Applying it to a limit such as $lim_{x \to 0} (sin x + x) / x^{2}$ — where only the numerator approaches zero — gives the wrong answer.
Ignoring one-sided behaviour. Checking only one side of a limit (e.g. only $x \to 0^{+}$ ) can miss a jump discontinuity. Always verify both left-hand and right-hand limits explicitly.
Confusing “the value at the point” with “the limit at the point”. For a removable hole, the limit exists and equals $L$ even though $f (a)$ may be undefined or equal some other value. They are logically independent.

Limit explorer

Squeeze the two approach markers toward the target and watch both sides converge on the limit value.

Example:

x \to 1 lim \frac{x ^{2} - 1}{x - 1} = 2

distance from x = 10.800

from the left f(0.20)1.2000

from the right f(1.80)2.8000

limit L2

There's a hole at x = 1 (the function is 0/0 there), yet the limit is a clean 2. Shrink the distance toward zero and both sides close in on L — that's what the limit means.

Worked example 1The difference quotient for a rate law

The concentration of a reactant is measured at two times: $[A] (t) = 1.00 - 0.040 t M$ (linear model, $t$ in s). Compute the average rate over $[5, 5 + Δ t]$ and then take the limit as $Δ t \to 0$ .

What is the instantaneous rate at $t = 5 s$ ?

Worked example 2Limit at infinite dilution — osmotic pressure

The van't Hoff equation for osmotic pressure is $Π = c RT$ for dilute solutions, but real solutions deviate. The true thermodynamic osmotic pressure satisfies $lim_{c \to 0} Π/ c = RT$ . Evaluate this limit at 25 °C ( $T = 298 K$ , $R = 8.314 J mo l^{- 1} K^{- 1}$ ).

Worked examples

Limits in chemical context — try each before revealing the solution.

Worked example 3Concentration approaching equilibrium

In a reversible first-order reaction $A ⇌ B$ with forward rate constant $k_{f} = 0.10 s^{- 1}$ and reverse $k_{r} = 0.040 s^{- 1}$ , the approach to equilibrium gives $[A] (t) = [A]_{eq} + ([A]_{0} - [A]_{eq}) e^{- (k_{f} + k_{r}) t}$ . Show that $lim_{t \to \infty} [A] (t) = [A]_{eq}$ .

Worked example 4One-sided limits and a discontinuous titration model

A simplified step-function model of a strong-acid–strong-base titration sets pH = 3 before the equivalence point and pH = 11 after. At exactly $V_{eq}$ , the model is undefined. Do the one-sided limits exist? Does the (two-sided) limit exist? Is the model continuous at $V_{eq}$ ?

Worked example 5L'Hôpital's rule on a kinetics expression

Show that $lim_{k_{2} \to k_{1}} \frac{e ^{- k_{1} t} - e ^{- k_{2} t}}{k _{2} - k _{1}}$ (which arises in consecutive first-order reactions when the two rate constants become equal) equals $t e^{- k_{1} t}$ .

Worked example 6Instantaneous rate from the difference quotient

A second-order reaction gives $[A] (t) = \frac{[ A ] _{0}}{1 + k [ A ] _{0} t}$ with $[A]_{0} = 0.50 M$ and $k = 0.020 L mo l^{- 1} s^{- 1}$ .

Use the difference-quotient definition to derive $d [A] / d t$ and evaluate the instantaneous rate at $t = 10 s$ .

Worked example 7Behaviour as t → ∞ in a consecutive reaction

In the consecutive scheme $A k_{1} B k_{2} C$ with $k_{2} \neq = k_{1}$ , the concentrations are $[A] = e^{- k_{1} t}$ , $[B] = \frac{k _{1}}{k _{2} - k _{1}} (e^{- k_{1} t} - e^{- k_{2} t})$ , and $[C] = 1 - [A] - [B]$ (with $[A]_{0} = 1$ ). Find all three limits as $t \to \infty$ .

Worked example 8L'Hôpital on a 0/0 kinetics ratio

Evaluate $lim_{t \to 0} \frac{1 - e ^{- k t}}{t}$ , which arises when estimating the initial rate from an integrated first-order law without using the derivative directly.

ChallengeLimit of [B](t) for A → B → C as k₂ → k₁

In the consecutive reaction $A k_{1} B k_{2} C$ with $[A]_{0} = 1$ , the intermediate is:

[B] (t) = \frac{k _{1}}{k _{2} - k _{1}} (e^{- k_{1} t} - e^{- k_{2} t})

This expression is indeterminate when $k_{2} = k_{1}$ . Use L'Hôpital's rule (treating $k_{2}$ as the variable) to find $lim_{k_{2} \to k_{1}} [B] (t)$ , and verify the physical limit as $t \to \infty$ of the result.

Check yourself

Four quick questions tying limits to chemical context.

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