Question

Discuss the continuity of the function f, where f is defined by

\[f(n) =   \begin{cases}     3,       & \quad \text{if } {0\leq x\leq 1}\\     4,  & \quad \text{if } {1<x<3} \\ 5,  & \quad \text{if } {3\leq x \leq 10}  \end{cases}\]

Answer 1

The given function is 
\(f(n) =   \begin{cases}     3,       & \quad \text{if } {0\leq x\leq 1}\\     4,  & \quad \text{if } {1<x<3} \\ 5,  & \quad \text{if } {3\leq x \leq 10}  \end{cases}\)
The given function f is defined at all the points of the interval [0,10]
Let c be a point on the interval [0,10]       

Case (I):
If 0≤c<1, then f(c) = 3 and \(\lim\limits_{x \to c}\) f(x) = \(\lim\limits_{x \to c}\) f(3) = 3
∴\(\lim\limits_{x \to c}\) f(x) = f(c)
Therefore, f is continuous in the interval [0,1).

Case (II):
If c=1, then f(3)=3
The left hand limit of f at x=1 is
\(\lim\limits_{x \to 1^-}\) f(x) =\(\lim\limits_{x \to 1^-}\)(3) = 3
The right hand limit of f at x=1 is,
\(\lim\limits_{x \to 1^+}\) f(x) =\(\lim\limits_{x \to 1^+}\)(4) = 4
It is observed that the left and right hand limit of f at x=1 do not coincide. 
Therefore, f is not continuous at x=1

Case(III):
If1<c<3, then f(c)=4 and
\(\lim\limits_{x \to c}\) f(x) = \(\lim\limits_{x \to c}\)(4) = 4
∴\(\lim\limits_{x \to c}\) f(x) = f(c)
Therefore, f is continuous at all points of the interval (1,3).

Case(IV):
If c=3, then f(c) = 5
The left hand limit of f at x=3 is,
\(\lim\limits_{x \to 3^-}\) f(x) = \(\lim\limits_{x \to 3^-}\)(4) = 4
The right hand limit of f at x=3 is,
\(\lim\limits_{x \to 3^+}\) f(x) =\(\lim\limits_{x \to 3^+}\)(5) = 5
It is observed that the left and right hand limits of f at x=3 do not coincide. 
Therefore, f is not continuous at x=3

Case(V):
If 3<c≤10, then f(c)=5 and \(\lim\limits_{x \to c}\) f(x) = \(\lim\limits_{x \to c}\) (5) = 5
\(\lim\limits_{x \to c}\) f(x) = f(c)
Therefore,f is continuous at all points of the interval (3,10].

Hence,f is not continuous at x=1 and x=3