Discuss the continuity of the function `f` , where `f` is defined by
`f(x)={{:(2x "," if x lt 0 ),(0"," if 0 le x le 1),(4x "," if x gt 1 ):}`

Explanation

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 lt 0`, then `f(c) = 2c`
`lim_(x->c)f(x)=lim_(x->c)(2x)=2c`
`lim_(x->c)f(x)=f(c)`
Therefore, f is continuous at all points x, such tha`t x lt 0`
Case II:If c = 0, then f(c) = f(0) = 0
The left hand \limit of f at x = 0 is,
`lim_(x->0)f(x)=lim_(x->0)(2x)=0=0`
The right hand \limit of f at x = 0 is,
`lim_(x->0)f(x)=lim_(x->0)(0)=0=0`
`lim_(x->0)f(x)=f(0)`
Therefore, f is continuous at x = 0
Case III :If ` 0 lt c lt 1`, then` f(x) = 0
`and `lim_(x->0)f(x)=lim_(x->c)(0)=0=0`
​`lim_(x->c)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,
`lim_(x->1)f(x)=lim_(x->1)(0)=0`
The right hand \limit of f at` x = 1`
`lim_(x->1)f(x)=lim_(x->1)(4xx1)=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
`lim_(x->c)f(x)=lim_(x->c)(4x)=4c`
`lim_(x->c)f(x)=f(c)`
Therefore, f is continuous at all points x, such that `x gt1`
Hence, f is not continuous only at x = 1