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Module 3 · L02 of 15 ~35 min +95 XP available

Limits

What happens to a function as its input gets closer and closer to a particular value? Sometimes the answer is obvious. Sometimes it is hidden behind a division by zero that needs to be untangled. The concept of a limit is the key that unlocks all of calculus — and in this lesson, you will learn how to find it.

Today's hook — Consider $f(x) = \frac{x^2 - 1}{x - 1}$. If you substitute $x = 1$, you get $\frac{0}{0}$, which is undefined. But what happens if you try values very close to 1, like $x = 0.9$, $x = 0.99$, or $x = 1.001$? What number do you think the function is approaching?

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Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

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Recall — your gut answer first

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Consider $f(x) = \frac{x^2 - 1}{x - 1}$. If you substitute $x = 1$, you get $\frac{0}{0}$, which is undefined. Without using algebra — what number do you think $f(x)$ approaches as $x$ gets closer and closer to 1? Try $x = 0.9$, $0.99$, $1.001$ in your head.

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The two moves

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There are only two core ideas in this lesson — and they unlock everything that follows in calculus. A limit describes where a function is heading, not where it actually lands. And a limit only exists when both sides agree.

Every limit problem in this module travels along one of two roads: direct substitution works for continuous functions, while algebraic simplification (factoring, cancelling) rescues you when substitution gives an indeterminate form like $\frac{0}{0}$.

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

Limit notation

$\lim_{x \to a} f(x) = L$ means $f(x)$ gets arbitrarily close to $L$ as $x$ approaches $a$. The value at $x = a$ does not matter.

One-sided limits

$\lim_{x \to a^-} f(x)$ (left) and $\lim_{x \to a^+} f(x)$ (right). The overall limit exists only if both one-sided limits exist and are equal.

Limit laws

Sum, product, quotient: $\lim[f(x) + g(x)] = \lim f(x) + \lim g(x)$, provided each individual limit exists.

What you will master

Know

Key facts

The meaning of limit notation $\lim_{x \to a} f(x)$
One-sided limits and the condition for existence of a limit
The limit laws (sum, product, quotient)
The difference between limits at a point and limits as $x \to \infty$

Understand

Concepts

What a limit means intuitively (approaching, not reaching)
Why limits matter for derivatives and rates of change
The connection between limits and continuity

Can do

Skills

Evaluate limits by direct substitution
Evaluate limits by factoring and cancelling
Evaluate limits from a graph
Evaluate one-sided limits for piecewise functions

Key terms

LimitThe value $L$ that $f(x)$ approaches as $x$ gets arbitrarily close to $a$: $\lim_{x \to a} f(x) = L$.

One-sided limitA limit approached from only one side: left-hand ($x \to a^-$) or right-hand ($x \to a^+$).

Left-hand limit$\lim_{x \to a^-} f(x)$: the value $f(x)$ approaches as $x$ approaches $a$ from below.

Right-hand limit$\lim_{x \to a^+} f(x)$: the value $f(x)$ approaches as $x$ approaches $a$ from above.

Limit lawsAlgebraic rules for splitting limits across sums, products, quotients and powers.

Continuity$f$ is continuous at $a$ if $\lim_{x \to a} f(x) = f(a)$ — the limit equals the function value.

DiscontinuityA point where $f$ is not continuous: jump, infinite, or removable.

Removable discontinuityA hole in the graph where $\lim_{x \to a} f(x)$ exists but $f(a)$ is undefined or different.

Limits — what's actually going on

core concept

The notation $\lim_{x \to a} f(x) = L$ means: as $x$ gets arbitrarily close to $a$ (from either side), the value of $f(x)$ gets arbitrarily close to $L$. Crucially, we do not care what happens exactly at $x = a$ — only what happens near it.

Example: A hole in the graph

Consider $f(x) = \frac{x^2 - 1}{x - 1}$. At $x = 1$, substituting gives $\frac{0}{0}$, which is undefined. But for every other value of $x$:

$$f(x) = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \quad \text{(provided } x \neq 1\text{)}$$

So as $x$ approaches 1, $f(x)$ approaches $1 + 1 = 2$. We write:

$$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2$$

The limit at x = 1 is 2, even though f(1) is undefined — the hole shows where the function does not exist.

Evaluating limits

Method 1: Direct substitution. If $f(x)$ is continuous at $x = a$, simply substitute:

$$\lim_{x \to 3} (2x + 5) = 2(3) + 5 = 11$$

Method 2: Factor and cancel. If substitution gives $\frac{0}{0}$, factor, cancel the common factor, then substitute:

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

Method 3: One-sided limits. For piecewise functions, check the left-hand and right-hand limits separately. The overall limit exists only if both one-sided limits exist and are equal.

Limit laws. Provided each individual limit exists:

Sum: $\lim[f(x) + g(x)] = \lim f(x) + \lim g(x)$
Product: $\lim[f(x) \cdot g(x)] = \lim f(x) \cdot \lim g(x)$
Quotient: $\lim\frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}$, provided $\lim g(x) \neq 0$
Power: $\lim[f(x)]^n = [\lim f(x)]^n$

Why this matters for engineering. When engineers design aircraft wings or bridge cables, they need to know how materials behave under stress as parameters approach extreme values. Limits let them predict behaviour at theoretical boundaries without ever reaching them — ensuring structures remain safe even under conditions that cannot be physically tested.

Limit notation: $\lim_{x \to a} f(x) = L$ means $f(x)$ approaches $L$ as $x$ approaches $a$ from either side; The value at $x = a$ is irrelevant — only what happens near $a$ matters

Pause — copy the limit notation $\lim_{x \to a} f(x) = L$ and the key idea that the value at $x = a$ is irrelevant — only the behaviour near $a$ matters into your book.

Quick check: True or false — $\lim_{x \to a} f(x)$ always equals $f(a)$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · SUBSTITUTION

Evaluate $\lim_{x \to 3} (x^2 + 2x)$ by substitution.

$f(x) = x^2 + 2x$ is a polynomial, so it is continuous everywhere.

Check continuity — polynomials are always continuous.

PROBLEM 2 · FACTORING

Evaluate $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ by factoring.

Substitution gives $\frac{4 - 4}{2 - 2} = \frac{0}{0}$ (indeterminate).

Direct substitution fails — we need to simplify first.

PROBLEM 3 · ONE-SIDED LIMITS

For the piecewise function $f(x) = \begin{cases} x + 1 & x < 2 \\ 5 & x = 2 \\ 7 - x & x > 2 \end{cases}$, find the left and right limits at $x = 2$ and determine if $\lim_{x \to 2} f(x)$ exists.

Left-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x + 1) = 2 + 1 = 3$

For $x < 2$, use the top branch $f(x) = x + 1$.

Quick check: Evaluate $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$.

Common errors · the 3 traps that cost marks

Trap 01

Thinking $\lim_{x \to a} f(x) = f(a)$ always

This is only true for continuous functions. A limit can exist even when $f(a)$ is undefined or different from the limit. For $f(x) = \frac{x^2 - 4}{x - 2}$, the limit at $x = 2$ is 4, but $f(2)$ is undefined. Always evaluate the limit by examining behaviour near $a$, not just at $a$.

Trap 02

Forgetting to check one-sided limits at piecewise boundaries

When a function changes its formula at a point (like $x = 2$ in a piecewise function), the left and right limits may differ. Students often check only one side and conclude the limit exists when it does not. Always check both sides at boundary points.

Trap 03

Canceling incorrectly when factoring

When you cancel $(x - a)$ from numerator and denominator, you are really dividing by $(x - a)$, which is only valid when $x \neq a$. The cancelled expression is equal to the original everywhere except at $x = a$ — which is exactly what you need for a limit. Write “for $x \neq a$” to show you understand this.

Odd one out: Three of these limits can be evaluated by direct substitution. Which one cannot?

Work mode · how are you completing this lesson?

Quick-fire practice · 5 reps

Find $\lim_{x \to 1} (3x + 2)$.

Find $\lim_{x \to -1} \frac{x^2 - 1}{x + 1}$.

For $f(x) = \begin{cases} x + 1 & x < 2 \\ 5 & x = 2 \\ 7 - x & x > 2 \end{cases}$, find the left and right limits at $x = 2$.

Use limit laws to find $\lim_{x \to 2} (x^2 + 3x - 1)$.

Explain why $\lim_{x \to 0} \frac{|x|}{x}$ does not exist.

Fill the blanks: drag each token into the matching blank.

direct substitution factor and cancel does not exist equal

For continuous functions, evaluate limits by ___. When substitution gives $\frac{0}{0}$, ___ the common factor. The overall limit exists only when the one-sided limits are ___; otherwise the limit ___.

Two truths and a lie: Identify the false statement about limits.

Revisit your thinking

Earlier you were asked about $f(x) = \frac{x^2 - 1}{x - 1}$ near $x = 1$.

Although $f(1)$ gives $\frac{0}{0}$ (undefined), we can factor the numerator: $x^2 - 1 = (x - 1)(x + 1)$. For all $x \neq 1$, this means $f(x) = x + 1$. So as $x$ approaches 1, $f(x)$ approaches $1 + 1 = 2$. The limit is 2, even though the function has a “hole” at $x = 1$.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 4

Q1. Evaluate $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$. Show working. 2 MARKS

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ApplyBand 4

Q2. For the piecewise function $f(x) = \begin{cases} 2x + 1 & x \leq 1 \\ 4 - x & x > 1 \end{cases}$, find $\lim_{x \to 1^-} f(x)$, $\lim_{x \to 1^+} f(x)$, and determine if $\lim_{x \to 1} f(x)$ exists. 3 MARKS

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AnalyseBand 5

Q3. Explain why the function $f(x) = \frac{x^2 - 4}{x - 2}$ has a removable discontinuity at $x = 2$. How could you redefine $f(2)$ to make the function continuous? 3 MARKS

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📖 Comprehensive answers (click to reveal)

Drill 1: $\lim_{x \to 1} (3x + 2) = 3(1) + 2 = 5$

Drill 2: $\lim_{x \to -1} \frac{(x-1)(x+1)}{x+1} = \lim_{x \to -1} (x-1) = -2$

Drill 3: Left: $\lim_{x \to 2^-} (x+1) = 3$. Right: $\lim_{x \to 2^+} (7-x) = 5$. Since $3 \neq 5$, $\lim_{x \to 2} f(x)$ does not exist.

Drill 4: By limit laws: $\lim_{x \to 2} (x^2 + 3x - 1) = 4 + 6 - 1 = 9$

Drill 5: Left: $\lim_{x \to 0^-} \frac{|x|}{x} = -1$. Right: $\lim_{x \to 0^+} \frac{|x|}{x} = 1$. Since $-1 \neq 1$, the limit does not exist.

Q1 (2 marks): $\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3} (x+3) = 6$ [2].

Q2 (3 marks): $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (2x+1) = 3$ [1]. $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4-x) = 3$ [1]. Since left = right = 3, $\lim_{x \to 1} f(x) = 3$ [1].

Q3 (3 marks): $f(x) = \frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2$ for $x \neq 2$ [1]. So $\lim_{x \to 2} f(x) = 4$, but $f(2)$ is undefined [1]. Redefine $f(2) = 4$ to make the function continuous [1].

Boss battle · The Approacher

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Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering limit questions. Lighter alternative to the boss.

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Module overview · Maths Advanced · Checkpoint 1