Question

Discuss the continuity of the function f, where f is defined by
f(x) = $\{\begin{array}{c}2x,ifx<0\\ 0,if0\le x\le 1\\ 4x,ifx>1\end{array}$

Answer 1

The given function is f(x) = $\{\begin{array}{c}2x,ifx<0\\ 0,if0\le x\le 1\\ 4x,ifx>1\end{array}$
The given function is defined at all points of the real line.
Let c be a point on the real line.
CAse I :
If c < 0, then f(c) = 2c
$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ (2x) = 2c
$\therefore$ $\underset{x\to c}{lim}$ f(x) = f(c)
Therefore, f is continuous at all points x, such that x < 0
Case II:
If c = 0, then f(c) = f(0) = 0
The left hand \limit of f at x = 0 is,
$\underset{x\to 0}{lim}$ f(x) = $\underset{x\to 0}{lim}$ (2x) = 2 x 0 = 0
The right hand \limit of f at x = 0 is,
$\underset{x\to 0}{lim}$ f(x) = $\underset{x\to 0}{lim}$ (0) = 0
$\therefore$ $\underset{x\to 0}{lim}$ f(x) = f(0)
Therefore, f is continuous at x = 0
Case III :
If 0 < c < 1, then f(x) = 0 and $\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ (0) = 0
$\therefore$ $\underset{x\to c}{lim}$ f(x) = f(c)
Therefore, f is continuous at all points of the interval (0, 1).
Case IV :
If c = 1, then f(c) = f(1) = 0
The left hand \limit of f at x = 1 is,
$\underset{x\to 1}{lim}$ f(x) = $\underset{x\to 1}{lim}$ (0) = 0
The right hand \limit of f at x = 1 is,
$\underset{x\to 1}{lim}$ f(x) = $\underset{x\to 1}{lim}$ (4x) = 4 x 1 = 4
it is observed that left and right hand \limit of f at x = 1 do not coincide
Therefore, f is not continuous at x = 1
Case V:
If c < 1, then f(c) = 4c and $\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ (4x) = 4c
$\therefore$ $\underset{x\to c}{lim}$ f(x) = f(c)
Therefore, f is continuous at all points x, such that x > 1
Hence, f is not continuous only at x = 1