Continuity at a Point and on an Interval

MATContinuity

A function $f$ is continuous at a point $x=a$ when its left-hand limit, right-hand limit and functional value all coincide. On an interval, continuity must hold at every interior point (and one-sidedly at endpoints).

ISC examines this through piecewise functions with unknown constants $k$ , $a$ , $b$ that you solve by equating limits to $f(a)$ , frequently using standard limits like $\dfrac{\sin x}{x}\to 1$ and $\dfrac{e^x-1}{x}\to 1$ .

Continuity at a point

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

f

is continuous at

x=a

iff all three quantities exist and are equal;

f(a)

must be defined.

Trigonometric standard limit

\lim_{x \to 0} \dfrac{\sin x}{x} = 1 \quad\text{and}\quad \lim_{x \to 0} \dfrac{1-\cos x}{x^2} = \dfrac{1}{2}

x

in radians; used to evaluate

0/0

forms such as

\dfrac{1-\cos 2x}{2x^2}

, which

\to 1

Exponential and logarithmic limits

\lim_{x \to 0} \dfrac{e^x - 1}{x} = 1, \qquad \lim_{x \to 0} \dfrac{\log(1+x)}{x} = 1

\log

is natural log; gives

\dfrac{e^{2x}-1}{x}\to 2

via

\dfrac{e^{2x}-1}{2x}\cdot 2

Rationalisation limit form

\lim_{x \to 0} \dfrac{\sqrt{1+kx}-\sqrt{1-kx}}{x} = k

Obtained by multiplying by the conjugate

\sqrt{1+kx}+\sqrt{1-kx}

; numerator becomes

2kx

and the denominator

\to 2

To find an unknown constant, set the relevant one-sided limit(s) equal to $f(a)$ and solve the resulting equation.
For continuity on $\mathbb{R}$ of a piecewise function, only the join points (where the formula changes) need checking; elsewhere each piece is built from continuous elementary functions.
At a join point you usually equate the limit of the left piece, the limit of the right piece and the assigned value, giving as many equations as unknowns.
For a removable discontinuity (a $0/0$ form), the correct value of the constant equals $\lim_{x\to a} f(x)$ computed from the non-trivial branch.
Always work in radians for trigonometric limits, and convert composite arguments, e.g. $1-\cos 2x = 2\sin^2 x$ .
Polynomials, $\sin x$ , $\cos x$ , $e^x$ are continuous everywhere; $\tan x$ , $\log x$ , $\dfrac{1}{x}$ are continuous only on their domains.
A function continuous on a closed interval $[a,b]$ is bounded and attains its maximum and minimum (a useful checking idea).

Where the marks go

Equating only one one-sided limit to $f(a)$ and forgetting to check the other side at a join point.
Using $\dfrac{\sin x}{x}\to 1$ with $x$ in degrees, or mismatching the argument, e.g. treating $\dfrac{1-\cos 2x}{2x^2}$ as $\dfrac12$ instead of its correct value $1$ .
Cancelling a factor before confirming the form is $0/0$ , or substituting $x=a$ directly into a piece that is not the one defining $f(a)$ .
Concluding continuity merely because $f(a)$ is defined, without verifying the limit equals $f(a)$ .

How the board asks it

Numericalstandard trigonometric limit and $f(a)$
For what value of $k$ is the function $f(x)=\dfrac{\sin 5x}{3x}$ for $x\neq 0$ and $f(0)=k$ continuous at $x=0$ ? Find $k$ .
Numericalequating one-sided limits at join points
Find the values of $a$ and $b$ so that $f(x)=\begin{cases} 5, & x\leq 2 \\ ax+b, & 2<x<10 \\ 21, & x\geq 10 \end{cases}$ is continuous on $\mathbb{R}$ .
Give reasonslimit must equal $f(a)$ ; removable discontinuity
Examine the continuity of $f(x)=\begin{cases} \dfrac{x^2-9}{x-3}, & x\neq 3 \\ 5, & x=3 \end{cases}$ at $x=3$ , and state, with reasons, whether the discontinuity is removable.
Derive / provecontinuity on an interval; modulus functions
Show that the function $f(x)=|x-2|+|x+1|$ is continuous for all $x\in\mathbb{R}$ .
Numericaltrigonometric limit at $x=\frac{\pi}{2}$
Find the value of $k$ for which $f(x)=\begin{cases} \dfrac{k\cos x}{\pi-2x}, & x\neq \frac{\pi}{2} \\ 3, & x=\frac{\pi}{2} \end{cases}$ is continuous at $x=\dfrac{\pi}{2}$ .

Practise this topic

NextDifferentiability and First Principles

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.