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CBC Curriculum Note: Limits and continuity form the
gateway topic of the Calculus strand in Senior Secondary Pure
Mathematics. KICD requires learners to evaluate limits using
algebraic techniques, understand continuity and its graphical
interpretation, and appreciate limits as the conceptual basis for
differentiation and integration — all of which are examined in
KCSE Paper 2 and form the entry point for university-level analysis
courses.
A limit describes what value a function f(x) gets arbitrarily
close to as x approaches some value a — without necessarily
caring what happens (or whether anything happens) exactly at
x = a. We write this as limx→a f(x) = L, read
"the limit of f(x) as x approaches a equals L."
This distinction — "approaching" versus "equalling" — is what
makes limits powerful. Many of the most important functions in
mathematics are undefined or behave strangely at a single point,
but are perfectly well-behaved everywhere nearby. Limits let us
describe that nearby behaviour precisely, which is exactly what
is needed to define the gradient of a curve at a point (a limit
of gradients of chords) or the area under a curve (a limit of
sums of rectangles).
Evaluating Limits by Direct Substitution
For most "ordinary" functions — polynomials, and combinations of
functions that are continuous at a — the limit as x→a is simply
found by substituting x = a directly. Problems only arise when
direct substitution produces an indeterminate form
such as 0/0 or ∞/∞, which signals that algebraic simplification is
needed first.
Worked Examples
Example 1 — Direct substitution limx→3 (x² + 2x − 1)
Step 1 The function is a polynomial, so it is continuous everywhere — substitute x = 3 directly.
Step 2 (3)² + 2(3) − 1 = 9 + 6 − 1
= 14
Example 2 — Indeterminate form 0/0 limx→2 (x² − 4)/(x − 2)
Step 1 Direct substitution gives 0/0 — an indeterminate form. We cannot conclude the limit is 0, 1, or undefined from this alone.
Step 2 Factorise the numerator: x² − 4 = (x−2)(x+2)
Step 3 Cancel the common factor (valid since x ≠ 2 while taking the limit): (x−2)(x+2)/(x−2) = x + 2
Step 4 Now substitute x = 2 into the simplified expression: 2 + 2
= 4
Example 3 — Limits at infinity limx→∞ (3x² + 2)/(x² − 1)
Step 1 Direct substitution gives ∞/∞ — indeterminate. Divide every term by the highest power of x in the denominator, x².
Step 2 = (3 + 2/x²)/(1 − 1/x²)
Step 3 As x→∞, 2/x² → 0 and 1/x² → 0, leaving (3+0)/(1−0)
= 3
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Recognising indeterminate forms: 0/0 and ∞/∞ are
the most common, but ∞−∞, 0×∞, 1^∞, 0⁰ and ∞⁰ are also
indeterminate. Each requires its own algebraic manipulation —
factorising, rationalising, dividing by the highest power, or
rewriting using logarithms — before the limit can be evaluated.
Sometimes a function approaches different values depending on
whether x approaches a from below (the left) or from above
(the right). One-sided limits let us describe each direction
separately — and a two-sided limit exists only when both
directions agree.
Worked Example
Example f(x) = |x|/x for x ≠ 0. Investigate limx→0 f(x)
Step 1 For x > 0, |x| = x, so f(x) = x/x = 1. Thus limx→0⁺ f(x) = 1
Step 2 For x < 0, |x| = −x, so f(x) = −x/x = −1. Thus limx→0⁻ f(x) = −1
Step 3 The left-hand and right-hand limits differ (−1 ≠ 1)
limx→0 f(x) does not exist
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"Does not exist" is a valid answer: Students
sometimes feel a limit question must have a numerical answer.
When one-sided limits disagree, or when a function oscillates
without settling (like sin(1/x) near 0), or grows without
bound, the correct and complete answer is that the limit
does not exist — write this explicitly.
Limit laws let us break a complicated limit into simpler pieces,
provided each piece's limit exists (and, for quotients, the
denominator's limit is non-zero). These laws are the formal
justification for "substituting term by term" that you used
intuitively in Example 1 above.
The Squeeze (Sandwich) Theorem
When a function f(x) is "trapped" between two other functions
g(x) and h(x) that both tend to the same limit L, then f(x) must
also tend to L — even if f(x) itself is too complicated to
evaluate directly. This is the standard tool for proving the
famous limit limx→0 sin x / x = 1.
Worked Example
Example — Squeeze theorem Evaluate limx→0 x² sin(1/x)
Step 1 sin(1/x) oscillates between −1 and 1 for all x ≠ 0, so −1 ≤ sin(1/x) ≤ 1
Step 2 Multiply through by x² (≥ 0): −x² ≤ x² sin(1/x) ≤ x²
Step 3 Both −x² and x² tend to 0 as x→0, so by the squeeze theorem the middle expression must also tend to 0
= 0
A small set of "famous" limits recur throughout calculus,
particularly when differentiating trigonometric and exponential
functions from first principles. Recognising these patterns —
including disguised versions of them — is a key A-Level skill.
Worked Example
Example — Disguised standard limit limx→0 sin(3x)/x
Step 1 Rewrite to expose the standard form: sin(3x)/x = 3 · sin(3x)/(3x)
Step 2 As x→0, 3x→0 too, so sin(3x)/(3x) → 1 by the standard limit
= 3 × 1 = 3
L'Hôpital's Rule
When a limit of a quotient gives the indeterminate form 0/0 or
∞/∞, L'Hôpital's Rule allows us to differentiate the numerator
and denominator separately (not as a quotient
rule!) and take the limit of the new ratio. If the new limit is
still indeterminate, the rule can be applied again.
Worked Example
Example — Repeated application limx→0 (x − sin x)/x³
Step 1 Direct substitution gives 0/0. Differentiate top and bottom: (1 − cos x)/(3x²) — still 0/0
Step 2 Differentiate again: (sin x)/(6x) — still 0/0
Step 3 Differentiate a third time: (cos x)/6
Step 4 Now substitute x = 0: cos(0)/6 = 1/6
= 1/6
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L'Hôpital applies only to indeterminate forms:
Always check that direct substitution genuinely gives 0/0 or
∞/∞ before differentiating top and bottom. If the limit is not
indeterminate, L'Hôpital's Rule does not apply and will give a
wrong answer.
Informally, a function is continuous at a point if its graph
has no break, jump, or hole there — you could draw it through
that point without lifting your pen. Formally, this requires
three conditions to all hold simultaneously.
The Three Conditions for Continuity
- f(a) is defined — the function actually has a value at x = a.
- limx→a f(x) exists — the left-hand and right-hand limits agree.
- limx→a f(x) = f(a) — the limiting value matches the actual function value.
If a function satisfies all three conditions at every point of an
interval, it is said to be continuous on that interval. Failing
any one of the three conditions produces a discontinuity, and the
way in which it fails determines the type of discontinuity.
Types of Discontinuity
Type
Description & Example
Removable
A "hole" — the limit exists but either f(a) is undefined or f(a) ≠ the limit e.g. f(x) = (x²−1)/(x−1) at x = 1
Jump
Left-hand and right-hand limits both exist but are unequal e.g. the floor function ⌊x⌋ at integer values
Infinite
The function grows without bound near a — a vertical asymptote e.g. f(x) = 1/x at x = 0
Oscillatory
The function oscillates infinitely often with no settling value e.g. f(x) = sin(1/x) at x = 0
Worked Example
Example Show that f(x) = (x² − 1)/(x − 1) has a removable discontinuity at x = 1, and state how to remove it.
Step 1 f(1) is undefined (0/0), so condition 1 fails — f is discontinuous at x = 1.
Step 2 For x ≠ 1, simplify: (x−1)(x+1)/(x−1) = x+1, so limx→1 f(x) = 1+1 = 2. The limit exists.
Step 3 Because the limit exists but f(1) is undefined, the discontinuity is removable — define f(1) = 2 to make f continuous.
Removable discontinuity at x = 1; redefine f(1) = 2
The Intermediate Value Theorem (IVT) is a direct consequence of
continuity: a continuous function cannot "skip" values. If it
takes one value at one point and a different value at another
point, it must take every value in between somewhere in the
interval.
The most common application of the IVT in Senior Secondary and
A-Level mathematics is proving that a root of an equation
f(x) = 0 exists within a given interval, by showing that f
changes sign over that interval.
Worked Example
Example Show that x³ − x − 2 = 0 has a root between x = 1 and x = 2
Step 1 Let f(x) = x³ − x − 2. Polynomials are continuous everywhere, so f is continuous on [1, 2].
Step 2 f(1) = 1 − 1 − 2 = −2
Step 3 f(2) = 8 − 2 − 2 = 4
Step 4 f(1) = −2 < 0 and f(2) = 4 > 0, so 0 lies between f(1) and f(2). By the IVT, there exists c ∈ (1,2) with f(c) = 0.
A root exists in (1, 2)
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The IVT proves existence, not value: The
theorem guarantees a root exists somewhere in the interval, but
does not tell you its exact value. Finding the actual value
requires further techniques such as bisection (a numerical
method directly motivated by the IVT) or algebraic methods.
Everything covered so far has relied on an intuitive idea of
"getting close to." The epsilon-delta (ε–δ) definition makes
this idea completely rigorous, removing any reliance on
intuition or motion-based language like "approaches."
In plain language: no matter how small a "tolerance band" ε you
draw around L on the y-axis, it is always possible to find a
corresponding "input window" δ around a on the x-axis, such that
every x inside that window (other than a itself) maps to a value
of f(x) inside the tolerance band. The challenge in a proof is
always to find an explicit δ in terms of ε.
Worked Example
Example Prove that limx→2 (3x − 1) = 5 using the ε–δ definition
Step 1 — Target We need: for every ε > 0, find δ > 0 such that 0 < |x−2| < δ ⟹ |(3x−1) − 5| < ε
Step 2 — Simplify the target expression |(3x−1) − 5| = |3x − 6| = 3|x − 2|
Step 3 — Choose δ We want 3|x−2| < ε, i.e. |x−2| < ε/3. So choose δ = ε/3.
Step 4 — Verify If 0 < |x−2| < δ = ε/3, then |(3x−1)−5| = 3|x−2| < 3·(ε/3) = ε ✔
δ = ε/3 works for every ε > 0, so limx→2(3x−1) = 5 is proven
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Why this matters: Every limit law, every
theorem about continuity, and the entire foundation of calculus
ultimately rests on this single definition. University courses
in real analysis spend significant time proving familiar
results (like the limit laws used earlier in this lesson) from
this ε–δ definition alone.
A sequence is a list of numbers a₁, a₂, a₃, ... indexed by the
natural numbers. The limit of a sequence asks the same question
as before — what value does aₙ approach as n becomes very
large? — but with n restricted to whole numbers rather than
moving continuously.
Common Series Convergence Tests
Test
Idea
Ratio test
Compute L = lim|aₙ₊₁/aₙ|. Converges if L < 1, diverges if L > 1, inconclusive if L = 1
Root test
Compute L = lim ⁿ√|aₙ|. Same conclusions as the ratio test
Integral test
For positive decreasing aₙ = f(n), Σaₙ converges iff ∫₁^∞ f(x)dx converges
Comparison test
If 0 ≤ aₙ ≤ bₙ and Σbₙ converges, then Σaₙ converges too
Worked Example
Example Determine whether aₙ = (2n+1)/(n+3) converges, and if so find its limit
Step 1 Divide numerator and denominator by n (the highest power present): aₙ = (2 + 1/n)/(1 + 3/n)
Step 2 As n→∞, 1/n→0 and 3/n→0
Step 3 aₙ → (2+0)/(1+0)
Converges to 2
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Connection to series: The geometric series
Σrⁿ converges to 1/(1−r) precisely when |r| < 1 — this
single fact, itself a limit of partial sums, underlies
applications from compound interest to the Sequences &
Series strand of the CBC syllabus.
A consolidated reference of every result covered in this lesson —
useful for revision before KCSE Paper 2 or university calculus
assessments.
Wrong limx→2 (x²−4)/(x−2) = 0/0 = "undefined, so the limit doesn't exist"
Correct Factorise and cancel first: the limit equals 4
0/0 means "indeterminate, needs more work" — not "the limit does not exist." Always simplify before concluding.
Wrong Applying L'Hôpital's Rule as a quotient rule: differentiate f/g as one expression
Correct Differentiate f and g separately, then form the new ratio f′/g′
L'Hôpital's Rule is not the quotient rule — mixing the two produces a completely incorrect result.
Wrong Concluding limx→0 f(x) = f(0) for every function
Correct This equality only holds if f is continuous at 0
For discontinuous functions, the limit and the function value at a point can differ — or one may not exist at all.
Wrong "The limit doesn't exist because the function isn't defined at that point"
Correct The limit can exist even when f(a) is undefined (removable discontinuity)
A limit only depends on values near a, never on the value (or absence of a value) exactly at a.
Wrong limx→∞ (3x²+2)/(x²−1), dividing only the numerator by x²
Correct Divide every term, top and bottom, by the same power of x
Dividing only part of the expression changes its value — the division must be applied consistently throughout.
Wrong Using the IVT to claim the root's exact value is known
Correct The IVT only guarantees a root exists in the interval
Finding the value itself requires further work — e.g. the bisection method or algebraic solving.