Find the point of discontinuity, if any, of the following function: \[f\left( x \right) = \begin{cases}\sin x - \cos x , & \text{ if } x \neq 0 \\ - 1 , & \text{ if } x = 0\end{cases}\]

#### 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.