Introduction To Continuity
We start with a very intuitive introduction to continuity. Consider the two graphs given in the figure below:
Our purpose is to analyse the behaviour of these functions around the region x = 1.
The obvious visual difference between the two graphs around x = 1 is that whereas the first graph passes uninterrupted (without a break) through x = 1, the second function suffers a break at x = 1 (there is a jump).
This visual difference, put into mathematical language, gives us the concept and definition of continuity. Mathematically, we say that the function \(f\left( x \right) = {\left( {x - 1} \right)^2}\) is continuous at x = 1 while \(f\left( x \right) = \left[ x \right]\) is discontinuous at x = 1
For \(f\left( x \right) = {\left( {x - 1} \right)^2},\)
\(LHL \left( {{\rm{at}}\,x = 1} \right) = \mathop {\lim }\limits_{x \to {1^ - }} {\left( {x - 1} \right)^2} = 0\)
and
\(RHL \left( {{\rm{at}}\,x = 1} \right) = \mathop {\lim }\limits_{x \to {1^ + }} {\left( {x - 1} \right)^2} = 0\)
and
\(f(1) = 0\)
\(\Rightarrow LHL = RHL = f(1)\)
For \(f\left( x \right) = \left[ x \right],\)
\({\rm{LHL }}\left( {{\rm{at\, }}x = 1} \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left[ x \right] = 0\)
and
\({\rm{RHL }}\left( {{\rm{at\, }}x = 1} \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left[ x \right] = 1\)
and
\(f(1) = 1\)
\(\Rightarrow \,\,\,\,\,{\rm{ }}\,{\rm{LHL}} \ne {\rm{RHL}}\,{\rm{ = }}\,\,f\left( 1 \right)\)
From the discussion above, try to see that for a function to be continuous at x = a, all the three quantities, namely, LHL, RHL and f (a) should be equal. In any other scenario, the function becomes discontinuous.
Discontinuities therefore arise in the following cases:
(a) One or more than one of the three quantities, LHL, RHL and f(a) is not defined . Lets consider some examples:
(i) \(\begin{align}f\left( x \right) = \frac{1}{x}\end{align}\) around x = 0.
\({\rm{LHL}}\,{\rm{ = }}\, - \infty ,{\rm{ RHL}}\,{\rm{ = }}\, + \infty ,\,f\left( 0 \right)\) is not defined. Therefore, \(\begin{align}f\left( x \right) = \frac{1}{x}\end{align}\) is discontinuous at x = 0 which is obvious from the graph:
(ii) \(\begin{align}f\left( x \right) = \left\{ {\frac{{{x^2} - 1}}{{x - 1}}{\rm{\,for\, }}x \ne 1} \right\}\end{align}\) around x = 1
\(LHL{\rm{ }} = {\rm{ }}RHL{\rm{ }} = {\rm{ }}2\) but \(f(1)\) is not defined. Therefore, this function’s graph has a hole at x = 1; it is discontinuous at x = 1:
(b) All the three quantities are defined, but any pair of them is unequal (or all three are unequal). Lets go over some examples again:
(i) \(f(x){\rm{ }} = {\rm{ }}[x] \text{around any integer }I\)
\(LHL{\rm{ }} = I-1,{\rm{ }}RHL{\rm{ }} = I,f\left( I \right){\rm{ \;}} = I\)
\(\Rightarrow {\rm{LHL}} \ne {\rm{RHL}}\,{\rm{ = }}\,f\left( I \right)\) so this function is discontinuous at all integers as
we already know.
(ii) \(f\left( x \right) = \left\{ x \right\}{\rm{ }}around{\rm{\; }}any{\rm{\; }}integer{\rm{\;}}I\)
\(LHL{\rm{ }} = 1,{\rm{ }}RHL{\rm{ }} = {\rm{ }}0{\rm{ }},f\left( I \right){\rm{ }} = {\rm{ }}0\)
\(\Rightarrow {\rm{LHL}} \ne {\rm{RHL}}\;{\rm{ = }}\;f\left( I \right)\) so this function is also discontinuous at all integers.
(iii) \(f\left( x \right) = \left\{ \begin{gathered}1,\,\,\,x \notin \mathbb{Z} \hfill \\0,\,\,\,x \in \mathbb{Z} \hfill \\\end{gathered} \right\}\) around any integer I
From the figure, we notice that at any integer \(I,{\rm{ }}LHL{\rm{ }} = {\rm{ }}1,{\rm{ }}RHL{\rm{ }} = 1,f\left( I \right){\rm{ }} = {\rm{ }}0\)
\( \Rightarrow {\rm{LHL}}\;{\rm{ = }}\;{\rm{RHL}} \ne f\left( I \right)\) so that this function is again discontinuous.
(iv) \(f\left( x \right) = \left\{ \begin{align}&\frac{{\left| x \right|}}{x},x \ne 0\\&0\,\,\,\,\,\,x = 0\end{align} \right\}{\rm{ around }}x = 0\)
At x = 0, we see that
\(LHL = -1,{\rm{ }}RHL{\rm{ }} = 1,f\left( 0 \right){\rm{ }} = {\rm{ }}0\)
\( \Rightarrow {\rm{LHL}} \ne {\rm{RHL}} \ne f\left( 0 \right)\) and this function is discontinuous.
To summarize, if we intend to evaluate the continuity of a function at x = a, which means that we want to determine whether f (x) will be continuous at x = a or not, we have to evaluate all the three quantities, LHL, RHL and f (a). If these three quantities are finite and equal, f (x) is continuous at x = a. In all other cases, it is discontinuous at x = a
\[\boxed{{\text{LHL}}\left( {{\text{at }}x = a} \right) = {\text{RHL}}\left( {{\text{at }}x = a} \right) = f\left( a \right):{\text{ for continuity at }}x = a}\]
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