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.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Consider the function $f(x) = \frac{x^2 - 1}{x - 1}$. 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?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: A stationary point is always a maximum or minimum.
Right: Stationary points can also be horizontal points of inflection where the concavity changes.
📚 Core Content
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.
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:
$$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2$$
If $f(x)$ is continuous at $x = a$, simply substitute:
$$\lim_{x \to 3} (2x + 5) = 2(3) + 5 = 11$$
If substitution gives $\frac{0}{0}$, factor the expression, cancel the common factor, and then substitute:
$$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2} = \lim_{x \to 2} (x + 2) = 4$$
For limits involving square roots, multiply numerator and denominator by the conjugate:
$$\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x}$$
Multiply top and bottom by $\sqrt{x + 1} + 1$:
$$= \lim_{x \to 0} \frac{(x + 1) - 1}{x(\sqrt{x + 1} + 1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x + 1} + 1)} = \lim_{x \to 0} \frac{1}{\sqrt{x + 1} + 1} = \frac{1}{2}$$
As $x$ grows without bound, the behaviour of a rational function is dominated by its highest powers. For example:
$$\lim_{x \to \infty} \frac{3x^2 + 2x}{5x^2 - 1} = \frac{3}{5}$$
This is because the $x^2$ terms dominate; the lower-power terms become negligible.
🧮 Worked Examples
🧪 Activities
1 $\lim_{x \to 2} (3x + 1)$
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2 $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$
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3 $\lim_{x \to 4} \frac{x^2 - 6x + 8}{x - 4}$
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4 $\lim_{x \to \infty} \frac{5x^2 + 3}{2x^2 - x}$
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1 For $f(x) = \frac{x^2 - 4}{x - 2}$, the value $f(2)$ is undefined but $\lim_{x \to 2} f(x) = 4$. Explain how this is possible.
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2 When finding $\lim_{x \to \infty} \frac{3x + 1}{x^2 + 5}$, a student divides top and bottom by $x^2$ and gets 0. Explain why the limit is 0 in practical terms.
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Answer in your workbook.
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 $x = 1$.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. Evaluate each of the following limits. Show all algebraic working. 4 MARKS
(a) $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$
(b) $\lim_{x \to \infty} \frac{4x^3 - 2x + 1}{2x^3 + 5x^2}$
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7. A student claims: "If $f(5) = 12$, then $\lim_{x \to 5} f(x)$ must equal 12." Is this claim always true? Provide a counter-example or a justification to support your answer. 3 MARKS
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8. Engineers use computer simulations to model how a bridge cable stretches as the load on it increases. The model predicts that as the load $L$ (in tonnes) approaches 1000 tonnes, the stretch $S$ (in cm) approaches 15 cm according to $S = \frac{15L}{L + 50}$. By evaluating an appropriate limit, verify the engineers' prediction and explain what it means physically. 3 MARKS
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Answer in your workbook.
1. $3(2) + 1 = 7$
2. $\lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1} (x + 1) = 2$
3. $\lim_{x \to 4} \frac{(x - 4)(x - 2)}{x - 4} = \lim_{x \to 4} (x - 2) = 2$
4. Divide by $x^2$: $\lim_{x \to \infty} \frac{5 + \frac{3}{x^2}}{2 - \frac{1}{x}} = \frac{5}{2}$
1. A limit considers the value that $f(x)$ approaches as $x$ gets close to 2, not the value at $x = 2$ itself. For $x \neq 2$, $f(x) = x + 2$, which approaches 4.
2. As $x \to \infty$, the denominator $x^2 + 5$ grows much faster than the numerator $3x + 1$. The fraction becomes arbitrarily small, so the limit is 0.
1. C — Substitute $x = 2$: $2^2 + 1 = 5$.
2. C — Factor: $\frac{(x-3)(x+3)}{x-3} = x+3$, which approaches $6$ as $x o 3$.
3. B — Divide numerator and denominator by $x^2$: $\frac{2 + \frac{1}{x^2}}{3 - \frac{1}{x^2}} o \frac{2}{3}$.
4. B — A limit describes the value that the function approaches as $x$ gets arbitrarily close to $a$.
5. B — Multiply by the conjugate: $\frac{(\sqrt{x+4}-2)(\sqrt{x+4}+2)}{x(\sqrt{x+4}+2)} = \frac{x}{x(\sqrt{x+4}+2)} o \frac{1}{4}$.
6. A — Divide by $x$: $\frac{5 + \frac{2}{x}}{1 - \frac{1}{x}} o 5$.
7. C — Factor: $\frac{(x-1)(x+1)}{x-1} = x+1$, which approaches $2$ as $x o 1$.
Q6 (4 marks): (a) $\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3} (x+3) = 6$ [2]. (b) Divide numerator and denominator by $x^3$: $\lim_{x \to \infty} \frac{4 - \frac{2}{x^2} + \frac{1}{x^3}}{2 + \frac{5}{x}}$ [1]. As $x \to \infty$, the fractional terms tend to 0, so the limit is $\frac{4}{2} = 2$ [1].
Q7 (3 marks): The claim is not always true [1]. A limit describes the value that $f(x)$ approaches as $x$ gets close to $a$, which need not equal $f(a)$ [1]. For example, let $f(x) = \frac{x^2 - 25}{x - 5}$ for $x \neq 5$ and $f(5) = 12$. Then $f(5) = 12$ but $\lim_{x \to 5} f(x) = \lim_{x \to 5} (x+5) = 10 \neq 12$ [1].
Q8 (3 marks): Evaluate $\lim_{L \to \infty} \frac{15L}{L + 50} = \lim_{L \to \infty} \frac{15}{1 + \frac{50}{L}} = 15$ cm [2]. This means that under extremely large loads the cable stretch approaches a maximum of 15 cm; it gets arbitrarily close to 15 cm but does not exceed it [1].
Limits
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