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प्रश्न
Find the points of discontinuity, if any, of the following functions: \[f\left( x \right) = \begin{cases}2x , & \text{ if } & x < 0 \\ 0 , & \text{ if } & 0 \leq x \leq 1 \\ 4x , & \text{ if } & x > 1\end{cases}\]
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उत्तर
The given function is \[f\left( x \right) = \begin{cases}2x , & \text{ if } & x < 0 \\ 0 , & \text{ if } & 0 \leq x \leq 1 \\ 4x , & \text{ if } & x > 1\end{cases}\]
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`
`lim_(x->c)=lim_(x->c)(c)=2c`
`∴lim_(N->oo)f(x)=f(c)`
Therefore, f is continuous at all points x, such that x < 0
Case II:
` " If " (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)=2xx0=0`
The right hand limit of f at x = 0 is,
`lim_(x->0)f(x)=lim_(x->0)(0)=0`
`∴lim_(x->0)(x)=f(0)`
Therefore, f is continuous at x = 0
Case III:
` " If " 0 < c< 1 " then " f(x) " and " lim_(x->c) f(x)=lim_(x->c)(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)f(1)=0`
The right hand limit of f at x = 1 is,
`lim_(x->1)f(x)=lim_(x->1)(4x)=4xx1=4`
It is observed that the left and right hand limits of f at x = 1 do not coincide.
Therefore, f is not continuous at x = 1
Case V:
` " If " c <, " then " f(c)=4c " and " lim_(x->c)f(4x)=4c`
`∴ lim_(x->c)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
