Examine the continuity of $f$,where $f$ is defined by $f(x) = \begin{cases} \sin x - \cos x, & \text{if } x \neq 0 \\ -1, & \text{if } x = 0 \end{cases}$

(A) The given function $f$ is defined as $f(x) = \begin{cases} \sin x - \cos x, & \text{if } x \neq 0 \\ -1, & \text{if } x = 0 \end{cases}$.
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 $f(c) = \sin c - \cos c$.
$\lim_{x \to c} f(x) = \lim_{x \to c} (\sin x - \cos x) = \sin c - \cos c$.
Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous at all points $x \neq 0$.
Case $II$: If $c = 0$,then $f(0) = -1$.
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0} (\sin x - \cos x) = \sin 0 - \cos 0 = 0 - 1 = -1$.
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0} (\sin x - \cos x) = \sin 0 - \cos 0 = 0 - 1 = -1$.
Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = -1$,$f$ is continuous at $x = 0$.
From the above observations,$f$ is continuous at every point of the real line.