Continuity of a function

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 A function \(f(x)\) is said to be continuous at \(x = a\) if \(f(x)\) converges to \(f(a)\) as \(x \to a\). Continuous functions are, in a sense, well-behaved and hence, easy to handle. 



Definition (Continuous function)

Let \(f(x)\) be a function defined on an interval \(I\) such that \(a \in I\). The function \(f(x)\) is said to be continuous at \(a\) if \[\lim_{x\to a}f(x) = f(a).\] \(f(x)\) is said to be a continuous function if it is continuous at every \(x\in I\).

Remark. According to the definition of limits, \(f(x)\) is continuous at \(x=a\) if the following condition is satisfied:

  • For any \(\varepsilon > 0\), there exists \(\delta > 0\) such that, for all \(x \in \mathrm{dom}f\), \(0 < |x - a| < \delta\) implies \(|f(x) - f(a)| < \varepsilon\).
□

Example. For the function \(f(x)\) defined by
\[f(x) = \left\{ \begin{array}{cc} 0 & (x < 0),\\ 1 & (x \geq 0), \end{array}\right.\]
\(\lim_{x \to 0}f(x)\) does not exist. Therefore it is not continuous at \(x=0\). □

Example. For the function \(f(x)\) defined by
\[ f(x) = \left\{ \begin{array}{cc} |x| & (x \neq 0),\\ -1 & (x = 0), \end{array}\right. \]
we have \(\lim_{x\to 0}f(x) = 0\). However, \(f(0) = -1 \neq \lim_{x\to 0}f(x)\). Therefore it is not continuous at \(x = 0\). □

Remark. If the function \(f(x)\) is defined on a closed interval \([a, b]\), we say it is continuous at \(x=a\) if
\[\lim_{x \to a + 0}f(x) = f(a),\]
[Note the right limit], and continuous at \(x = b\) if
\[\lim_{x \to b-0}f(x) = f(b)\]
[Note the left limit]. □

Theorem (Properties of continuous functions)

Let \(f(x)\) and \(g(x)\) be functions that are continuous at \(x = a\). Then the following functions are also continuous at \(x = a\).
  1. \(kf(x) + lg(x)\) where \(k, l\) are constants,
  2. \(f(x)g(x)\),
  3. \(\frac{f(x)}{g(x)}\) provided that \(g(a) \neq 0\).
Proof. Trivial from the definition of continuous functions and the properties of limits. ■

Example. Any polynomial functions are continuous at all \(a\in\mathbb{R}\). □

Theorem (Continuity of a composite function)

Let \(f(x)\) be a function defined on an interval including \(x=a\) such that \(b = f(a)\). Let \(g(x)\) be a function defined on an interval including \(x=b\). If \(f(x)\) is continuous at \(x=a\) and \(g(x)\) is continuous at \(x=b\), then \((g\circ f)(x)\) is continuous at \(x=a\).  
Proof. Trivial from the definition of continuity and limit of a composite function. ■

We give the following theorems without proof, but they show some of the important properties of continuous functions. Note that both theorems require the function to be defined on a closed interval.

Theorem (Intermediate Value Theorem)

Let \(f(x)\) be a continuous function defined on a closed interval \([a,b]\) such that \(f(a) \neq f(b)\). Then, for any \(l\) between \(f(a)\) and \(f(b)\), there exists a \(c\in[a,b]\) such that \(f(c) = l\).
See also: Intermediate Value Theorem (Wikipedia)

Theorem (Extreme Value Theorem)

A continuous function \(f(x)\) defined on a closed interval \([a, b]\) has a maximum value and a minimum value, each at least once.
See also: Extreme Value Theorem (Wikipedia)

Example. Consider the function
\[f(x) = \frac{1}{1 - x^2}\]
defined on \((-1, 1)\). This is a continuous function. However since \(\lim_{x\to 1-0}f(x) = +\infty\) and \(\lim_{x\to -1+0}f(x) = +\infty\) so it has neither a maximum nor a minimum. □


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