Neighbourhoods and the meaning of a limit – Limits of Polynomial, Rational and Irrational Functions (Senior 4 Mathematics) | Ganzaa

Limits of Polynomial, Rational and Irrational Functions

Lessons

Senior 4 · Mathematics

Limits of Polynomial, Rational and Irrational Functions

4 lessons

Lesson 1 of 4

Neighbourhoods and the meaning of a limit

25 min

A value you cannot calculate, but can approach

The function $f(x) = \frac{x^2 - 1}{x - 1}$ is not defined at $x = 1$ : substituting gives $\frac{0}{0}$ . Yet its values near 1 are perfectly ordinary:

$x$	0.9	0.99	0.999	1	1.001	1.01	1.1
$f (x)$	1.9	1.99	1.999	undefined	2.001	2.01	2.1

As $x$ gets closer to 1, from either side, $f (x)$ gets closer to 2. We say that the limit of $f (x)$ as $x$ tends to 1 is 2. Limits describe what a function does near a point, whatever happens at it. They are the foundation of calculus: Unit 8 defines speed and the derivative as limits.

Definition

Neighbourhood of a real number $a$ : an open interval $(a - \delta, a + \delta) = \{x : |x - a| \lt \delta\}$ with $\delta \gt 0$ . A punctured neighbourhood leaves out $a$ itself: $\{x : 0 \lt |x - a| \lt \delta\}$ .

A variable $x$ tends to $a$ (written $x \to a$ ) when $x$ takes values, different from $a$ , that eventually lie in every neighbourhood of $a$ , however small.

Example: $(2.9, 3.1)$ is the neighbourhood of 3 with $\delta = 0.1$ ; it is the solution set of $|x - 3| \lt 0.1$ (Unit 5).

Definition

Limit of a function (intuitive): $\displaystyle\lim_{x \to a} f(x) = L$ means that $f (x)$ can be made as close to $L$ as we like by taking $x$ close enough to $a$ , with $x \ne a$ .

Limit (precise, the " $\varepsilon$ – $\delta$ " definition): $\displaystyle\lim_{x \to a} f(x) = L$ if for every $\varepsilon \gt 0$ there is a $\delta \gt 0$ such that

0 \lt |x - a| \lt \delta \;\Longrightarrow\; |f(x) - L| \lt \varepsilon .

In words: whatever tolerance $\varepsilon$ is demanded around $L$ , there is a punctured neighbourhood of $a$ on which $f$ stays within it.

The graph of $f(x) = \frac{x^2 - 1}{x - 1}$ is the line $y = x + 1$ with a hole at $(1, 2)$ . For a tolerance $\varepsilon$ around 2 (orange band), every $x \ne 1$ in the blue band of half-width $\delta$ gives $f (x)$ inside the orange band. Here $\delta = \varepsilon$ works.

Worked example 1: proving a limit with $\varepsilon$ and $\delta$

Prove that $\displaystyle\lim_{x \to 2} (3x - 1) = 5$ .

Solution Let $\varepsilon \gt 0$ . We need $|(3x - 1) - 5| \lt \varepsilon$ , that is $3|x - 2| \lt \varepsilon$ , that is $|x - 2| \lt \frac{\varepsilon}{3}$ . So choose $\delta = \frac{\varepsilon}{3}$ : if $0 \lt |x - 2| \lt \delta$ , then $|(3x - 1) - 5| = 3|x - 2| \lt 3\delta = \varepsilon$ . ∎

For example, to keep $3 x - 1$ within $0.03$ of 5, it is enough to keep $x$ within $0.01$ of 2.

Important

Operations on limits. If $\displaystyle\lim_{x \to a} f(x) = L$ and $\displaystyle\lim_{x \to a} g(x) = M$ , then

$\lim (f \pm g) = L \pm M$ , $\lim (kf) = kL$ , $\lim (fg) = LM$ ;
$\lim \frac{f}{g} = \frac{L}{M}$ provided $M \ne 0$ ;
$\lim [f(x)]^n = L^n$ , and $\lim \sqrt[n]{f(x)} = \sqrt[n]{L}$ (with $L \ge 0$ when $n$ is even).

Consequences: for a polynomial, $\displaystyle\lim_{x \to a} P(x) = P(a)$ ; for a rational function, $\displaystyle\lim_{x \to a} \frac{P(x)}{Q(x)} = \frac{P(a)}{Q(a)}$ when $Q(a) \ne 0$ . In these cases, substitute.

Worked example 2: limits by substitution

Find (a) $\displaystyle\lim_{x \to 2} (x^3 - 4x + 1)$ (b) $\displaystyle\lim_{x \to -1} \frac{2x^2 + 3}{x - 4}$ (c) $\displaystyle\lim_{x \to 3} \sqrt{x^2 + 7}$ .

Solution

A polynomial: $8 - 8 + 1 = 1$ .
The denominator at $- 1$ is $-5 \ne 0$ : $\frac{2 + 3}{-5} = -1$ .
$\lim (x^2 + 7) = 16 \gt 0$ , so the limit is $\sqrt{16} = 4$ .

Warning

Substitution is a result of the limit laws, valid only when the denominator's limit is not 0 and the expression is defined around $a$ . If substitution gives $\frac{0}{0}$ , the limit may still exist (it was 2 in the opening example): that is an indeterminate form, treated in Lesson 3. Never write " $\frac{0}{0} = 1$ " or " $\frac{0}{0} = 0$ ".

Important

Squeeze (sandwich) theorem. If $g(x) \le f(x) \le h(x)$ for all $x$ in a punctured neighbourhood of $a$ , and $\displaystyle\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$ , then $\displaystyle\lim_{x \to a} f(x) = L$ .

Worked example 3: a function trapped between two parabolas

Find $\displaystyle\lim_{x \to 0} x^2 \sin\frac{1}{x}$ .

Solution The limit laws do not apply, because $\sin\frac{1}{x}$ has no limit at 0 (it keeps swinging between $- 1$ and $1$ ). But $-1 \le \sin\frac{1}{x} \le 1$ , and multiplying by $x^2 \ge 0$ :

-x^2 \le x^2 \sin\frac{1}{x} \le x^2 \quad (x \ne 0).

Both $- x^{2}$ and $x^{2}$ tend to 0, so by the squeeze theorem $\displaystyle\lim_{x \to 0} x^2 \sin\frac{1}{x} = 0$ .

$y = x^2 \sin\frac{1}{x}$ (blue) is trapped between $y = x^{2}$ and $y = - x^{2}$ (dashed), which both tend to 0, so it is squeezed to 0 as $x \to 0$ .

Tip

To find a limit, try substitution first. If it gives a number, that is the answer. If it gives $\frac{\text{non-zero}}{0}$ , expect an infinite limit (Lesson 2). If it gives $\frac{0}{0}$ , simplify (Lesson 3). If an oscillating factor such as $\sin\frac{1}{x}$ appears, think of the squeeze theorem.

3Blue1Brown: Limits, L'Hôpital's rule, and epsilon delta definitions (about 18 min). Watch the first part, on how a limit is defined by "for every ε there is a δ"; the part on L'Hôpital's rule belongs to Unit 8.

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Practice

Questions 7 and 8 use the precise definition and the squeeze theorem.

Write the neighbourhood of $- 2$ with radius $0.5$ as an interval and as an inequality with an absolute value.
Complete a table of values of $f(x) = \frac{x^2 - 4}{x - 2}$ for $x = 1.9, 1.99, 2.01, 2.1$ , and guess $\displaystyle\lim_{x \to 2} f(x)$ .
Find $\displaystyle\lim_{x \to -2} (x^4 - 3x^2 + x)$ .
Find $\displaystyle\lim_{x \to 1} \frac{x^2 + 3x}{2x + 1}$ .
Find $\displaystyle\lim_{x \to 4} \sqrt{2x + 1} \cdot (x - 1)$ .
Given $\displaystyle\lim_{x \to a} f(x) = 3$ and $\displaystyle\lim_{x \to a} g(x) = -2$ , find $\displaystyle\lim_{x \to a} \frac{f(x)^2 - g(x)}{f(x) + g(x)}$ .
Prove with $\varepsilon$ and $\delta$ that $\displaystyle\lim_{x \to -1} (5 - 2x) = 7$ .
For all $x$ , $3 - x^2 \le f(x) \le 3 + x^2$ . Find $\displaystyle\lim_{x \to 0} f(x)$ , naming the theorem you use.

Show answersHide answers

$(- 2.5, - 1.5)$ , that is $|x + 2| \lt 0.5$ .
$3.9, 3.99, 4.01, 4.1$ (since $f (x) = x + 2$ for $x \ne 2$ ); the limit is 4.
$16 - 12 - 2 = 2$ .
$\frac{4}{3}$ .
$\sqrt{9} \times 3 = 9$ .
$\frac{9 - (-2)}{3 + (-2)} = 11$ .
$|(5 - 2x) - 7| = 2|x + 1| \lt \varepsilon$ when $|x + 1| \lt \frac{\varepsilon}{2}$ ; take $\delta = \frac{\varepsilon}{2}$ .
Both bounds tend to 3, so by the squeeze theorem the limit is 3.

Quick check

Which inequality describes the neighbourhood of 5 with radius 0.2?

Key points

A neighbourhood of $a$ is $|x - a| \lt \delta$ ; $\lim_{x \to a} f(x)$ depends only on values near $a$ , never on $f (a)$ .
$\lim_{x \to a} f(x) = L$ : for every $\varepsilon \gt 0$ there is $\delta \gt 0$ with $0 \lt |x - a| \lt \delta \Rightarrow |f(x) - L| \lt \varepsilon$ .
Limit laws give substitution for polynomials, and for rational and irrational functions where the denominator is not 0 and the root is defined.
Squeeze theorem: a function trapped between two functions with the same limit has that limit too.

Sources and credits

Written by Ganzaa: the explanations, the graphs, the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
Image: Portrait of Augustin Louis Cauchy, lithograph by Louis Grégoire and Deneux, Smithsonian Libraries, Wikimedia Commons, public domain.
Video: Limits, L'Hôpital's rule, and epsilon delta definitions, 3Blue1Brown, embedded from YouTube.

1A value you cannot calculate, but can approach

A value you cannot calculate, but can approach