Indeterminate forms – 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 3 of 4

Indeterminate forms

25 min

Speed at one instant

A stone dropped from a tower falls $s (t) = 5 t^{2}$ metres in $t$ seconds (taking $g \approx 10\,\mathrm{m\,s^{-2}}$ ). Its average speed between the times $t$ and 2 is $\frac{s(t) - s(2)}{t - 2} = \frac{5t^2 - 20}{t - 2}$ . To get the speed at $t = 2$ , let $t \to 2$ . Substitution gives $\frac{0}{0}$ , which tells you nothing yet. But $\frac{5t^2 - 20}{t - 2} = \frac{5(t - 2)(t + 2)}{t - 2} = 5(t + 2)$ for $t \ne 2$ , which tends to $20$ . So the stone's speed at 2 s is $20\,\mathrm{m\,s^{-1}}$ . This lesson collects the algebra that turns such "undecided" limits into numbers.

Definition

Indeterminate form: a limit whose type, read from the limits of its parts, does not decide its value. The four you will meet are

\frac{0}{0}, \qquad \frac{\infty}{\infty}, \qquad \infty - \infty, \qquad 0 \times \infty .

Example: as $x \to 0$ , each of $\frac{x}{x}$ , $\frac{x^2}{x}$ and $\frac{x}{x^3}$ has the form $\frac{0}{0}$ , but their limits are $1$ , $0$ and $\pm\infty$ (no single value). The form alone decides nothing; you must transform the expression.

Important

Methods

Form	Typical expression	Method
$\frac{0}{0}$	rational function at a root $a$ of both numerator and denominator	factorise; cancel the common factor $x - a$ (factor theorem)
$\frac{0}{0}$	contains $\sqrt{A} - B$	multiply top and bottom by the conjugate $\sqrt{A} + B$
$\frac{\infty}{\infty}$	rational or irrational, $x \to \pm\infty$	divide by the highest power of $x$ (Lesson 2)
$\infty - \infty$	difference of fractions, or of roots	common denominator; or conjugate
$0 \times \infty$	product	rewrite as a single quotient

Worked example 1: factorise and cancel

Find (a) $\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 5x + 6}$ (b) $\displaystyle\lim_{x \to 1} \frac{x^3 - 1}{x^2 - 1}$ .

Solution Both give $\frac{0}{0}$ , so by the factor theorem $x - 3$ (or $x - 1$ ) divides the top and the bottom.

$\frac{(x - 3)(x + 3)}{(x - 3)(x - 2)} = \frac{x + 3}{x - 2}$ for $x \ne 3$ , which tends to $\frac{6}{1} = 6$ .
$\frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 1)} = \frac{x^2 + x + 1}{x + 1} \to \frac{3}{2}$ .

The $\frac{0}{0}$ at $x = 3$ is only a hole at $(3, 6)$ : the limit there is 6. At $x = 2$ the form is $\frac{-5}{0}$ , which gives an infinite limit and a vertical asymptote.

Worked example 2: multiply by the conjugate

Find (a) $\displaystyle\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$ (b) $\displaystyle\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}$ .

Solution Use $(\sqrt{A} - B)(\sqrt{A} + B) = A - B^2$ .

$\frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{(x - 4)(\sqrt{x} + 2)} = \frac{x - 4}{(x - 4)(\sqrt{x} + 2)} = \frac{1}{\sqrt{x} + 2} \to \frac{1}{4}$ .
$\frac{(x + 9) - 9}{x(\sqrt{x + 9} + 3)} = \frac{1}{\sqrt{x + 9} + 3} \to \frac{1}{6}$ .

Worked example 3: $\infty - \infty$

Find (a) $\displaystyle\lim_{x \to +\infty} \left(\sqrt{x^2 + 4x} - x\right)$ (b) $\displaystyle\lim_{x \to 2} \left(\frac{1}{x - 2} - \frac{4}{x^2 - 4}\right)$ .

Solution

Multiply and divide by the conjugate: $\sqrt{x^2 + 4x} - x = \frac{(x^2 + 4x) - x^2}{\sqrt{x^2 + 4x} + x} = \frac{4x}{x\left(\sqrt{1 + \frac{4}{x}} + 1\right)} = \frac{4}{\sqrt{1 + \frac{4}{x}} + 1} \to \frac{4}{2} = 2 .$
Common denominator: $\frac{(x + 2) - 4}{(x - 2)(x + 2)} = \frac{x - 2}{(x - 2)(x + 2)} = \frac{1}{x + 2} \to \frac{1}{4}$ .

Worked example 4: $0 \times \infty$ and $\frac{\infty}{\infty}$

Find (a) $\displaystyle\lim_{x \to +\infty} \frac{1}{x} \cdot \frac{2x^2 + 1}{x + 3}$ (b) $\displaystyle\lim_{x \to +\infty} \frac{\sqrt{9x^2 + 1}}{x + 5}$ .

Solution

The first factor tends to 0, the second to $+\infty$ . Combine into one quotient: $\frac{2x^2 + 1}{x^2 + 3x} \to 2$ (equal degrees).
$\frac{\infty}{\infty}$ . For $x \gt 0$ , $\sqrt{9x^2 + 1} = x\sqrt{9 + \frac{1}{x^2}}$ , so the quotient is $\frac{\sqrt{9 + \frac{1}{x^2}}}{1 + \frac{5}{x}} \to 3$ .

Warning

Cancelling $x - 3$ is allowed only because $x \ne 3$ while $x \to 3$ . The functions $\frac{x^2 - 9}{x^2 - 5x + 6}$ and $\frac{x + 3}{x - 2}$ are not equal at $x = 3$ (the first is undefined there), but they have the same limit. Also, $\infty - \infty$ is not 0: in Worked example 3(a) it was 2.

Tip

With a $\frac{0}{0}$ form in a rational function, you already know a factor: if substituting $a$ makes the top and the bottom zero, both are divisible by $x - a$ . Use synthetic division if the factorisation is not obvious.

The Organic Chemistry Tutor: Limits of rational functions, fractions and square roots (about 10 min). Watch how each 0 ÷ 0 limit is unlocked by factorising, by a common denominator or by a conjugate.

✦

Practice

Name the indeterminate form before you start each one.

$\displaystyle\lim_{x \to -2} \frac{x^2 + 5x + 6}{x^2 - 4}$
$\displaystyle\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$
$\displaystyle\lim_{x \to 1} \frac{x^3 - 3x + 2}{x^2 - 2x + 1}$
$\displaystyle\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}$
$\displaystyle\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x}$
$\displaystyle\lim_{x \to +\infty} \left(\sqrt{x^2 + x + 1} - x\right)$
$\displaystyle\lim_{x \to 3} \left(\frac{1}{x - 3} - \frac{6}{x^2 - 9}\right)$
A ball's height is $h (t) = 30 t - 5 t^{2}$ metres. Find $\displaystyle\lim_{t \to 1} \frac{h(t) - h(1)}{t - 1}$ and say what it means.

Show answersHide answers

$\frac{(x + 2)(x + 3)}{(x + 2)(x - 2)} = \frac{x + 3}{x - 2} \to \frac{1}{-4} = -\frac{1}{4}$ .
$x^2 + 2x + 4 \to 12$ .
$x^{3} - 3 x + 2 = (x - 1)^{2} (x + 2)$ , so the quotient is $x + 2 \to 3$ .
$x - 9 = (\sqrt{x} - 3)(\sqrt{x} + 3)$ , so the limit is $\sqrt{9} + 3 = 6$ .
Conjugate: $\frac{2x}{x(\sqrt{1 + x} + \sqrt{1 - x})} \to \frac{2}{2} = 1$ .
$\frac{x + 1}{\sqrt{x^2 + x + 1} + x} \to \frac{1}{2}$ .
$\frac{x + 3 - 6}{(x - 3)(x + 3)} = \frac{1}{x + 3} \to \frac{1}{6}$ .
$\frac{30t - 5t^2 - 25}{t - 1} = \frac{-5(t - 1)(t - 5)}{t - 1} = -5(t - 5) \to 20$ : the ball rises at $20\,\mathrm{m\,s^{-1}}$ at $t = 1$ .

Quick check

Match each limit with its indeterminate form.

Tap one on the left, then its partner on the right.

Key points

$\frac{0}{0}$ , $\frac{\infty}{\infty}$ , $\infty - \infty$ and $0 \times \infty$ are indeterminate: the answer depends on the expression, so transform it.
Rational $\frac{0}{0}$ at $a$ : factor out $x - a$ top and bottom and cancel. Roots: multiply by the conjugate.
Differences of fractions: common denominator. Products: rewrite as one quotient.
The limit of an average speed as the time interval shrinks is the speed at an instant (Unit 8).

Sources and credits

Written by Ganzaa: the explanations, the graph, 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.
Video: Limits of Rational Functions - Fractions and Square Roots, The Organic Chemistry Tutor, embedded from YouTube.

1Speed at one instant

Speed at one instant