# Limits and Continuity

> IB Mathematics: Analysis and Approaches HL · IB AA HL Calculus
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u5-limits-and-continuity/

Limits describe the behavior of functions near a point, and are the foundational concept for all calculus. This topic connects function properties to differentiation and integration, and introduces continuity as a key requirement for core calculus results.

**Prerequisites:** [Function properties and notation](https://www.owlsprep.com/study/ib-math-aa-hl-u2-functions-and-notation/); [Trigonometric identities](https://www.owlsprep.com/study/ib-math-aa-hl-u3-trigonometric-identities/)

## Learning objectives

- Evaluate limits of algebraic, trigonometric and rational functions using multiple methods
- Determine continuity of a function at a point and over an interval
- Classify different types of discontinuities
- Apply the Intermediate Value Theorem for continuous functions

## Definition and Basic Limit Evaluation

**Two-Sided Limit** — 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 $L$.

*Notation:* \lim_{x \to a} f(x) = 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 $\lim_{x \to 2} \frac{x^2 - 3x + 2}{x - 2}$

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. $$\lim_{x \to 2} (x-1) = 2 - 1 = 1$$

> **tip**
>
> Remember that $\frac{0}{0}$ is always indeterminate, but $\frac{c}{0}$ where $c \neq 0$ means the limit tends to positive or negative infinity.

## Standard Trigonometric Limits

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

- $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$
- $\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = 0$

**Worked example:** Evaluate $\lim_{x \to 0} \frac{\sin 3x}{x}$

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

## Continuity and Discontinuity Types

**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 $f(x) = \begin{cases} x^2 + 1 & x < 1 \\ 3 & x =1 \\ 2x & x>1 \end{cases}$ is continuous at $x=1$.

1. Check the three conditions for continuity:
2. 1. Is $f(1)$ defined? Yes, $f(1) = 3$.
3. 2. 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. 3. 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.

## Intermediate Value Theorem

**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.

## Common pitfalls

- **Wrong:** Assuming that if $f(a)$ is undefined, the limit as $x \to a$ does not exist
  - Why it fails: Many functions have removable discontinuities where the function is undefined but the limit exists
  - Correct: Always factor and simplify first to check if the limit exists after canceling common terms
- **Wrong:** Evaluating $\lim_{x \to 0} \frac{\sin kx}{x} = 1$ directly without adjusting
  - Why it fails: The standard limit only applies when the argument of sine matches the denominator
  - Correct: Rewrite to get $\frac{\sin kx}{x} = k \cdot \frac{\sin kx}{kx}$, so the limit equals $k$
- **Wrong:** Claiming a function is discontinuous at a point not in its domain
  - Why it fails: Continuity is only defined for points in the function's domain
  - Correct: Only discuss continuity at points that are part of the function's domain
- **Wrong:** Using the Intermediate Value Theorem on a non-continuous function
  - Why it fails: The IVT only holds for continuous functions; discontinuous functions can jump over values without crossing them
  - Correct: Always confirm the function is continuous on $[a,b]$ before applying IVT in a proof
- **Wrong:** Forgetting a two-sided limit only exists if left and right limits are equal
  - Why it fails: For piecewise functions, one-sided limits often exist but differ, so the overall limit does not exist
  - Correct: Always calculate both one-sided limits at points where the function definition changes

## 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 $x=a$ | 1. $f(a)$ defined; 2. Limit exists; 3. Limit = $f(a)$ | All three conditions required |
| Standard trig limits | $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} =1$, $\lim_{\theta \to 0} \frac{1-\cos \theta}{\theta}=0$ | 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 |

## 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.

- [L'Hôpital's rule (HL only)](https://www.owlsprep.com/study/ib-math-aa-hl-u5-l-h-pital-s-rule/)
- [Definition of the derivative](https://www.owlsprep.com/study/ib-math-aa-hl-u5-definition-of-the-derivative/)
- [Differentiation rules](https://www.owlsprep.com/study/ib-math-aa-hl-u5-differentiation-rules/)

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