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Question
Find the point of discontinuity, if any, of the following function: \[f\left( x \right) = \begin{cases}\sin x - \cos x , & \text{ if } x \neq 0 \\ - 1 , & \text{ if } x = 0\end{cases}\]
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Solution
The given function f is \[f\left( x \right) = \begin{cases}\sin x - \cos x , & \text{ if } x \neq 0 \\ - 1 , & \text{ if } x = 0\end{cases}\]
It is evident that f is defined at all points of the real line.
Let c be a real number.
Case I:
`" if c ≠ 0 , then " f ( c) = sin c - cos c `
`lim_(x → c) f ( x) = lim_ ( x→c ) ( sin x - cos x ) = sin c - cos c `
`∴ lim _ (x →c) f ( x) = f ( c) `
Therefore, f is continuous at all points x, such that x ≠ 0
Case II:
if c = 0 , then f (0) = - 1
`lim _ (x →0^-) f(x) = lim _ (x →0^-)(sin x - cos x ) = sin 0 - cos 0 = 0- 1 =- 1`
`lim _ (x →0^ +) f (x) = lim _ (x →0)(sin x - cos x ) = sin 0 - cos 0 = 0 - 1 = - 1`
`∴ lim _ (x →0^-) f (x) = lim _ (x →0^+) f (x)= f(0) `
Therefore, f is continuous at x = 0
From the above observations, it can be concluded that f is continuous at every point of the real line.
Thus, f is a continuous function.
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