
How do you find the discontinuity of a piecewise function?
Answer
447.3k+ views
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
\[\mathop {\lim }\limits_{x \to a} f(x) \ne f(a)\].
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
\[\mathop {\lim }\limits_{x \to a} f(x) \ne f(a)\].
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Express the following as a fraction and simplify a class 7 maths CBSE

The length and width of a rectangle are in ratio of class 7 maths CBSE

The ratio of the income to the expenditure of a family class 7 maths CBSE

How do you write 025 million in scientific notatio class 7 maths CBSE

How do you convert 295 meters per second to kilometers class 7 maths CBSE

Trending doubts
Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What is a transformer Explain the principle construction class 12 physics CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

How much time does it take to bleed after eating p class 12 biology CBSE
