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प्रश्न
Find the points of discontinuity, if any, of the following functions: \[f\left( x \right) = \begin{cases}- 2 , & \text{ if }& x \leq - 1 \\ 2x , & \text{ if } & - 1 < x < 1 \\ 2 , & \text{ if } & x \geq 1\end{cases}\]
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उत्तर
The given function f is \[f\left( x \right) = \begin{cases}- 2 , & \text{ if }& x \leq - 1 \\ 2x , & \text{ if } & - 1 < x < 1 \\ 2 , & \text{ if } & x \geq 1\end{cases}\]
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 < -1 " then " f(c)= -2 " and " lim_(x->c)(x)=lim_(x->c)(-2)=-2`
`∴lim_(x->c)f(x)=f(c)`
Therefore, f is continuous at all points x, such that x < −1
Case II:
` " If c =1 then " f(c)=f(-1)=-2`
The left hand limit of f at x = −1 is,
`lim_(x->-1)f(x)=lim_(x->-1)f(-2)=-2`
The right hand limit of f at x = −1 is,
`lim_(x->-1)f(x)=lim_(x->-1)f(2x)=2xx(-1)=-2`
`∴lim_(x->-1)f(x)=f(-1)`
Therefore, f is continuous at x = −1
Case III:
` " if -1 < c < 1,then " f(c)=2c`
`lim_(x->c)f(x)=lim_(x->c)f(2x)=2c`
`∴lim_(x->c)f(x)=f(c)`
Therefore, f is continuous at all points of the interval (−1, 1).
Case IV:
` " if c = 1,then " f(c)=f(1)=2xx1=2`
The left hand limit of f at x = 1 is,
`lim_(x->1)f(x)=lim_(x->1)2=2`
The right hand limit of f at x = 1 is,
`lim_(x->1)f(x)=lim_(x->1)2=2`
`∴ lim_(x->1)f(x)=lim_(x->1)f(c)`
Therefore, f is continuous at x = 2
Case V:
` " If c > 1 , then f(c)=2 and " lim_(x->1)f (x)=lim_(x->1)(2)=2 `
`lim_(x->c)f(x)=f(c)`
Therefore, f is continuous at all points x, such that x > 1
Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.
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