Examine the continuity of f, where f is defined by

Solution:

The given function $f$ is $f(x)= \begin{cases}\sin x-\cos x, & \text { if } x \neq 0 \\ -1 & \text { if } x=0\end{cases}$

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

If $c \neq 0$, then $f(c)=\sin c-\cos c$

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(\sin x-\cos x)=\sin c-\cos c$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Therefore, f is continuous at all points x, such that x ≠ 0

Case II:

If $c=0$, then $f(0)=-1$

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1$

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1$

$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$

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.