Question

How do you find the discontinuity of a piecewise function?

Answer 1

Hint: Piecewise function is a function that behaves differently based on the input ‘x’ values. We can solve the given problem by taking an example. We know that a function is said to be discontinuous at ‘a’ then \[\mathop {\lim }\limits_{x \to a} f(x) \ne f(a)\]. We can also say that if the limit does not exist then we can say that it is discontinuity.Here we need to find both the left hand limit and right hand limit.

Complete step by step answer:
Let’s take an example.Consider,
\[f(x) = \left\{
{x^2}{\text{ }}if{\text{ }}x < 1 \\
x{\text{ }}if{\text{ 1}} \leqslant x < 2 \\
2x - 1{\text{ }}if{\text{ 2}} \leqslant x \\
\right.\]
Thus we have taken a piecewise function in which the function defined at ‘x’ values 1 and 2.
Let’s check that if ‘f’ is continuous or discontinuous at \[x = 1\].
\[\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} {x^2} = {1^2} = 1\]
(Take the function which is defined at \[x < 1\])
\[\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} x = 1\]
(Take the function which is defined at \[1 \leqslant x\])
Since both limits give the same answer,
\[\mathop {\lim }\limits_{x \to 1} f(x) = 1{\text{ }} - - - - - (1)\].
\[f(1) = 1{\text{ }} - - - (2)\]
(Take the function which is defined at \[x = 1\])
From equation (1) and (2) we have,
\[\mathop {\lim }\limits_{x \to a} f(x) = f(a)\].
Hence at \[x = 1\] the given function is not discontinuous.
Let’s check that if ‘f’ is continuous or discontinuous at \[x = 2\].
\[\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} x = 2\]
(Take the function which is defined at \[x < 2\])
\[\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} (2x - 1) = (2(2) - 1) = 3\]
(Take the function which is defined at \[2 \leqslant x\])
We can see that both give different values hence, \[\mathop {\lim }\limits_{x \to 2} f(x)\] does not exist.

Hence, there is a jump discontinuity at \[x = 2\].

Note:We know that if both right hand limit and left hand limit exist, then only the limit exists. If the limit does not exist then no need to find f(a) value. We can say that it is discontinuous. We have three different kinds of discontinuity. In the given problem we have the left hand side limit is not equal to the right hand side limit. Hence it is jump discontinuity. That is
\[\mathop {\lim }\limits_{x \to {1^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {1^ + }} f(x)\]
If both the one sided limits are equal and it is not equal to f(x) at x=a then it is a removable discontinuity. That is
\[\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x)\] and
\[\mathop {\lim }\limits_{x \to a} f(x) \ne f(a)\].