Limits

Mathematics M2 (Algebra and Calculus) · Senior Secondary (HKDSE) · 7 min read · Updated: 24 Aug 2026

M2 Chapter: Limits - Your Comprehensive Study Notes

Hey everyone! Welcome to the foundational topic of Calculus: Limits. Don't worry if the word sounds a bit abstract. The goal of these notes is to make the concept of limits super clear and easy to understand. Limits are the primary building block for everything else in Calculus, including differentiation and integration. So, let's get started and build a strong foundation together!

1. What on Earth is a Limit? (The Intuitive Idea)

Imagine you're walking towards a wall. You take steps that are half the remaining distance. You walk halfway... then half of the remaining way... then half of that remaining way... and so on. You get incredibly close to the wall, but you never technically touch it.

In mathematics, a limit is the value that a function approaches as the input (usually \(x\)) gets closer and closer to some number.

The key idea is that we do not care what the function's value is exactly at the number. We only care about what value it is heading towards.

The Notation

We write the limit of a function like this:

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

Let's break that down:

lim is short for "limit".
\(x \to a\) means "as \(x\) approaches the value \(a\)".
\(f(x)\) is our function.
\(L\) is the value the function is approaching.

So, the whole expression reads: "The limit of the function \(f(x)\) as \(x\) approaches \(a\) is \(L\)."

One-Sided Limits and Existence of a Limit

A limit can be approached from two directions:

Left-hand limit: \(\lim_{x \to a^-} f(x)\) is the value \(f(x)\) approaches as \(x\) gets closer to \(a\) from values strictly smaller than \(a\) (the left side).
Right-hand limit: \(\lim_{x \to a^+} f(x)\) is the value \(f(x)\) approaches as \(x\) gets closer to \(a\) from values strictly greater than \(a\) (the right side).

Condition for Limit Existence: The overall two-sided limit \(\lim_{x \to a} f(x) = L\) exists if and only if both one-sided limits exist and are equal:

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

A Crucial Point: The Limit vs. The Actual Value

Sometimes, the limit at a point \(a\) is the same as \(f(a)\). But not always! This is a super important concept.

Example: Consider the function \(f(x) = \frac{x^2 - 4}{x - 2}\).

What happens at \(x = 2\)? If you plug it in, you get \(\frac{4-4}{2-2} = \frac{0}{0}\), which is undefined! There is a "hole" in the graph at \(x = 2\).

But what value does the function approach as \(x\) gets really close to \(2\)?

Let's try values near \(2\):
\(f(1.9) = 3.9\)
\(f(1.99) = 3.99\)
\(f(2.01) = 4.01\)
\(f(2.1) = 4.1\)

See? The closer \(x\) gets to \(2\), the closer \(f(x)\) gets to \(4\). So, we say:

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

Even though \(f(2)\) is undefined, the limit as \(x\) approaches \(2\) exists and is equal to \(4\).

Key Takeaway: Section 1

A limit is the target value a function is heading towards as its input gets closer to a certain number. It does not matter if the function ever actually reaches that value.

2. How to Find Limits: The Toolkit

We cannot just plug in nearby numbers every time. Here are the algebraic techniques you'll use in HKDSE M2.

Method 1: Direct Substitution (The Easy Way)

For polynomials and well-behaved functions defined at \(x = a\), the first thing you should try is substituting \(x = a\) directly.

Example 1: Find \(\lim_{x \to 3} (2x^2 - 5x + 1)\)

\(2(3)^2 - 5(3) + 1 = 2(9) - 15 + 1 = 18 - 15 + 1 = 4\)

So, \(\lim_{x \to 3} (2x^2 - 5x + 1) = 4\).

Method 2: When Direct Substitution Fails (The Indeterminate Form \(\frac{0}{0}\))

If direct substitution gives you an indeterminate form like \(\frac{0}{0}\), it does not mean the limit does not exist! It means you need to simplify the expression algebraically.

Technique A: Factorisation and Cancellation

Example: Find \(\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\)

Try Direct Substitution: We get \(\frac{2^2 - 4}{2 - 2} = \frac{0}{0}\).
Factor: \(x^2 - 4 = (x-2)(x+2)\).
Rewrite and Cancel:
\(\lim_{x \to 2} \frac{(x-2)(x+2)}{x-2}\)
Since \(x \to 2\), \(x \neq 2\), so \(x - 2 \neq 0\), meaning we can cancel the factor \((x-2)\):
\(\lim_{x \to 2} (x+2)\)
Substitute: \(2 + 2 = 4\). The limit is \(4\).

Technique B: Rationalisation (Conjugate Multiplication)

Use this technique when you see a square root and obtain \(\frac{0}{0}\).

Example: Find \(\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x}\)

Try Direct Substitution: \(\frac{\sqrt{0+1} - 1}{0} = \frac{0}{0}\).
Multiply Numerator and Denominator by the Conjugate \(\sqrt{x+1} + 1\):
\(\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} \times \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1}\)
Simplify: Using \((a-b)(a+b) = a^2 - b^2\):
\(\lim_{x \to 0} \frac{(\sqrt{x+1})^2 - 1^2}{x(\sqrt{x+1} + 1)} = \lim_{x \to 0} \frac{(x+1) - 1}{x(\sqrt{x+1} + 1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x+1} + 1)}\)
Cancel \(x\):
\(\lim_{x \to 0} \frac{1}{\sqrt{x+1} + 1}\)
Direct Substitution: \(\frac{1}{\sqrt{0+1} + 1} = \frac{1}{1+1} = \frac{1}{2}\).

3. Theorems on Limits

Limit theorems allow us to break down complicated expressions. The HKDSE syllabus requires using these rules without proof.

Let \(\lim_{x \to a} f(x) = L\) and \(\lim_{x \to a} g(x) = M\).

Sum/Difference Rule: \(\lim_{x \to a} [f(x) \pm g(x)] = L \pm M\)
Scalar Multiplication Rule: \(\lim_{x \to a} [k \cdot f(x)] = k \cdot L\) (where \(k\) is a constant)
Product Rule: \(\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M\)
Quotient Rule: \(\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}\) (provided \(M \neq 0\))
Power Rule: \(\lim_{x \to a} [f(x)]^n = L^n\) (for positive integers \(n\), and roots where defined)

4. Limits at Infinity

What happens to a function when \(x\) grows without bound (\(x \to \infty\) or \(x \to -\infty\))? We evaluate limits at infinity.

The fundamental theorem is:

\(\lim_{x \to \infty} \frac{1}{x^n} = 0\) and \(\lim_{x \to -\infty} \frac{1}{x^n} = 0\) (for any rational number \(n > 0\)).

Evaluating Limits of Rational Functions at Infinity

Standard Method: Divide every term in both the numerator and denominator by the highest power of \(x\) present in the denominator.

Example: Find \(\lim_{x \to \infty} \frac{4x^2 - 3x + 5}{2x^2 + 7x - 1}\)

The highest power in the denominator is \(x^2\).
Divide each term by \(x^2\):
\(\lim_{x \to \infty} \frac{\frac{4x^2}{x^2} - \frac{3x}{x^2} + \frac{5}{x^2}}{\frac{2x^2}{x^2} + \frac{7x}{x^2} - \frac{1}{x^2}} = \lim_{x \to \infty} \frac{4 - \frac{3}{x} + \frac{5}{x^2}}{2 + \frac{7}{x} - \frac{1}{x^2}}\)
Apply the rule \(\lim_{x \to \infty} \frac{c}{x^n} = 0\):
\(\frac{4 - 0 + 0}{2 + 0 - 0} = \frac{4}{2} = 2\)

5. Special Limits and the Constant \(e\)

HKDSE Exam Alert: These special limits are NOT provided on the HKDSE M2 formula sheet. You must memorize them and recognize how to apply them.

1. The Definition of \(e\) via Limits

The mathematical constant \(e \approx 2.71828\) is defined as:

\(\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e\)

Equivalently, letting \(x = \frac{1}{n}\), as \(n \to \infty\), \(x \to 0\):

\(\lim_{x \to 0} (1 + x)^{\frac{1}{x}} = e\)

2. Special Trigonometric Limit

\(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\) (where \(\theta\) is in radians)

Example: Find \(\lim_{x \to 0} \frac{\sin(5x)}{2x}\)

\(\lim_{x \to 0} \frac{\sin(5x)}{2x} = \lim_{x \to 0} \left(\frac{\sin(5x)}{5x} \cdot \frac{5}{2}\right) = 1 \cdot \frac{5}{2} = \frac{5}{2}\)

3. Special Exponential Limit

\(\lim_{x \to 0} \frac{e^x - 1}{x} = 1\)

Example: Find \(\lim_{x \to 0} \frac{e^{3x} - 1}{x}\)

\(\lim_{x \to 0} \frac{e^{3x} - 1}{x} = \lim_{x \to 0} \left(\frac{e^{3x} - 1}{3x} \cdot 3\right) = 1 \cdot 3 = 3\)

Chapter Summary: Master Strategy

Limits as \(x \to a\):
- First test Direct Substitution.
- If you get \(\frac{0}{0}\), factorise & cancel, rationalise using conjugates, or reshape expressions to use \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) or \(\lim_{x \to 0} \frac{e^x-1}{x} = 1\).
Limits as \(x \to \pm\infty\): Divide numerator and denominator by the highest power of \(x\) in the denominator and apply \(\lim_{x \to \infty} \frac{1}{x^n} = 0\).
Constant \(e\): Remember \(\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e\) and \(\lim_{x \to 0} (1+x)^{1/x} = e\).

References

Mathematics Education - Curriculum Documents — Education Bureau (EDB) / Curriculum Development Council (CDC), Hong Kong SAR Government
Ongoing Renewal of the Senior Secondary Curriculum (Mathematics M1/M2 Framework) — Education Bureau (EDB), Hong Kong SAR Government

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the intuitive definition of a limit in calculus?

A limit is the target value that a function approaches as its input gets closer and closer to a particular number, regardless of whether it actually reaches that value.

What is the necessary and sufficient condition for a two-sided limit to exist at a point?

The two-sided limit \(\lim_{x \to a} f(x) = L\) exists if and only if both the left-hand limit \(\lim_{x \to a^-} f(x)\) and the right-hand limit \(\lim_{x \to a^+} f(x)\) exist and are equal to \(L\).

Does the limit of \(f(x)\) as \(x \to a\) always equal the actual value \(f(a)\)?

No, the limit \(\lim_{x \to a} f(x)\) describes the value \(f(x)\) approaches and can exist even when \(f(a)\) is undefined or different from the limit.

What is the first step you should take when evaluating a limit of a well-behaved function as \(x \to a\)?

Perform direct substitution by plugging \(x = a\) directly into the function.

How do you evaluate an indeterminate limit of the form \(\frac{0}{0}\) containing polynomial expressions?

Factorise the numerator and denominator, cancel out the common factor causing the zero, and then use direct substitution.

How do you evaluate an indeterminate limit of the form \(\frac{0}{0}\) that contains square root terms?

Multiply the numerator and denominator by the conjugate of the radical expression to rationalise it, cancel the common factor, and then evaluate by direct substitution.

What are the primary algebraic limit laws for sums, products, and quotients?

If \(\lim f(x) = L\) and \(\lim g(x) = M\), then \(\lim [f(x) \pm g(x)] = L \pm M\), \(\lim [f(x) \cdot g(x)] = L \cdot M\), and \(\lim \frac{f(x)}{g(x)} = \frac{L}{M}\) (provided \(M \neq 0\)).

What is the standard method for evaluating the limit of a rational function at infinity (\(x \to \pm\infty\))?

Divide every term in both the numerator and denominator by the highest power of \(x\) present in the denominator, then apply \(\lim_{x \to \pm\infty} \frac{1}{x^n} = 0\).

What are the two standard limit definitions of Euler's number \(e\)?

\(\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e\) and \(\lim_{x \to 0} (1 + x)^{\frac{1}{x}} = e\).

What is the standard trigonometric limit involving the sine function as angle approaches 0?

\(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\), where \(\theta\) must be measured in radians.

What is the special exponential limit involving \(e^x\) as \(x \to 0\)?

\(\lim_{x \to 0} \frac{e^x - 1}{x} = 1\).

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