Limits and Continuity

IB Mathematics: Analysis and Approaches HL· Topic 5.1· 25 min read

1. Definition and Basic Limit Evaluation★★☆☆☆⏱ 8 min

📘 Definition

Two-Sided Limit

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

A function $f (x)$ has a limit $L$ as $x$ approaches $a$ if and only if the left-hand limit as $x \to a^{-}$ equals the right-hand limit as $x \to a^{+}$ , both equal to $L$ .

Example:

For $f (x) = \frac{x ^{2} - 1}{x - 1}$ , $lim_{x \to 1} f (x) = 2$

Limits can be evaluated using direct substitution for continuous functions, or factoring for rational functions with removable discontinuities. For limits at infinity, divide the numerator and denominator by the highest power of $x$ in the denominator.

📐 Worked Example

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

1
First test direct substitution: substituting $x = 2$ gives $\frac{0}{0}$ , an indeterminate form, so we factor the numerator:
2
$x^2 - 3x + 2 = (x-1)(x-2)$
3
Cancel the common $(x - 2)$ term: this is valid because we only consider values of $x$ near 2, not equal to 2, when evaluating the limit:
4
$\lim_{x \to 2} \frac{(x-1)(x-2)}{x-2} = \lim_{x \to 2} (x-1)$
5
Evaluate the simplified limit via direct substitution:
6
$\lim_{x \to 2} (x-1) = 2 - 1 = 1$

2. Standard Trigonometric Limits★★★☆☆⏱ 7 min

Two standard limits for trigonometric functions are frequently used in IB exams, and should be memorized:

$lim_{θ \to 0} \frac{s i n θ}{θ} = 1$
$lim_{θ \to 0} \frac{1 - c o s θ}{θ} = 0$

📐 Worked Example

Evaluate $lim_{x \to 0} \frac{s i n 3 x}{x}$

1
Rewrite the expression to match the standard form $lim_{θ \to 0} \frac{s i n θ}{θ}$ :
2
$\frac{\sin 3x}{x} = 3 \cdot \frac{\sin 3x}{3x}$
3
Let $θ = 3 x$ , so as $x \to 0$ , $θ \to 0$ . Substitute into the limit:
4
$\lim_{x \to 0} 3 \cdot \frac{\sin 3x}{3x} = 3 \cdot \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 3 \cdot 1 = 3$

3. Continuity and Discontinuity Types★★★☆☆⏱ 6 min

📘 Definition

Continuity at a Point

A function $f (x)$ is continuous at $x = a$ if and only if three conditions are met: 1) $f (a)$ is defined, 2) $lim_{x \to a} f (x)$ exists, 3) $lim_{x \to a} f (x) = f (a)$ .

Example:

$f (x) = x^{2}$ is continuous at $x = 0$ , since $lim_{x \to 0} x^{2} = 0 = f (0)$

Discontinuities are classified as removable or non-removable. Non-removable discontinuities include jump and infinite discontinuities. A function is continuous over an interval if it is continuous at every point in the interval.

📐 Worked Example

Determine if the piecewise function $f (x) = ⎩ ⎨ ⎧ x^{2} + 1 3 2 x x < 1 x = 1 x > 1$ is continuous at $x = 1$ .

1
Check the three conditions for continuity:
2
1. Is $f (1)$ defined? Yes, $f (1) = 3$ .
3
1. Does the limit exist? Find left and right hand limits:
4
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 +1) = 1 + 1 = 2$
5
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 2x = 2(1) = 2$
6
Left and right limits are equal, so $lim_{x \to 1} f (x) = 2$ exists.
7
1. Does the limit equal $f (1)$ ? $lim_{x \to 1} f (x) = 2 \neq = 3 = f (1)$ . So the function is not continuous at $x = 1$ . This is a removable discontinuity, since the limit exists.

4. Intermediate Value Theorem★★★★☆⏱ 4 min

📘 Definition

Intermediate Value Theorem (IVT)

If $f (x)$ is continuous on the closed interval $[a, b]$ , and $N$ is any number between $f (a)$ and $f (b)$ , then there exists at least one number $c$ in $(a, b)$ such that $f (c) = N$ . A common corollary is that if $f (a)$ and $f (b)$ have opposite signs, there is at least one root $c$ in $(a, b)$ where $f (c) = 0$ .

📐 Worked Example

Show that $f (x) = x^{3} - x - 1$ has a root between $x = 1$ and $x = 2$ .

1
First, $f (x)$ is a polynomial, so it is continuous on all real numbers, hence continuous on $[1, 2]$ .
2
Calculate $f (1)$ and $f (2)$ :
3
$f(1) = 1^3 - 1 - 1 = -1 < 0$
4
$f(2) = 8 - 2 - 1 = 5 > 0$
5
By IVT, since $0$ is between $- 1$ and $5$ , there exists a $c \in (1, 2)$ such that $f (c) = 0$ . Hence, a root exists on the interval.

5. Common Pitfalls

Wrong move:

Assuming that if $f (a)$ is undefined, the limit as $x \to a$ does not exist

Why:

Many functions have removable discontinuities where the function is undefined but the limit exists

Correct move:

Always factor and simplify first to check if the limit exists after canceling common terms

Wrong move:

Evaluating $lim_{x \to 0} \frac{s i n k x}{x} = 1$ directly without adjusting

Why:

The standard limit only applies when the argument of sine matches the denominator

Correct move:

Rewrite to get $\frac{s i n k x}{x} = k \cdot \frac{s i n k x}{k x}$ , so the limit equals $k$

Wrong move:

Claiming a function is discontinuous at a point not in its domain

Why:

Continuity is only defined for points in the function's domain

Correct move:

Only discuss continuity at points that are part of the function's domain

Wrong move:

Using the Intermediate Value Theorem on a non-continuous function

Why:

The IVT only holds for continuous functions; discontinuous functions can jump over values without crossing them

Correct move:

Always confirm the function is continuous on $[a, b]$ before applying IVT in a proof

Wrong move:

Forgetting a two-sided limit only exists if left and right limits are equal

Why:

For piecewise functions, one-sided limits often exist but differ, so the overall limit does not exist

Correct move:

Always calculate both one-sided limits at points where the function definition changes

6. Quick Reference Cheatsheet

Concept	Key Result	Notes
Two-sided limit exists	$lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = L$	Does not depend on $f (a)$
Continuity at $x = a$	$f (a)$ defined; 2. Limit exists; 3. Limit = $f (a)$	All three conditions required
Standard trig limits	$lim_{θ \to 0} \frac{s i n θ}{θ} = 1$ , $lim_{θ \to 0} \frac{1 - c o s θ}{θ} = 0$	Memorize for exam
Removable discontinuity	Limit exists, does not equal $f (a)$	Can be fixed by redefining $f (a)$
Non-removable discontinuity	Limit does not exist (jump/infinite)	Cannot be fixed with redefinition
Intermediate Value Theorem	Continuous on $[a, b]$ , $N$ between $f (a), f (b)$ → $f (c) = N$ for $c \in (a, b)$	Used to prove root existence

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · Paper 1
Evaluate limit of rational function
2022 · Paper 1
Prove continuity of piecewise function
2021 · Paper 1
Identify discontinuity type

Going deeper

SyllabusOfficial IB AA HL Calculus SyllabusOfficial topic reference

What's Next

Limits and continuity are the foundational building blocks for all of calculus. Every result in differentiation and integration relies on the formal definition of a limit, so mastering this topic will make all subsequent calculus topics much easier. Next, you will use limits to define the derivative from first principles, which connects this topic directly to differentiation rules for all function types. Continuity is also a key requirement for the Fundamental Theorem of Calculus, which links differentiation and integration, so understanding continuity is critical for solving integration problems correctly. The Intermediate Value Theorem is also frequently tested in proof questions on IB HL exams.