Examine the continuity off where f is defined by
$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\sin x - \cos x,}&{if}&{x \ne 0}\\{ - 1,}&{if}&{x = 0}\end{array}} \right.$ }}

We have

$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 0 - h\atop\scriptstyle h \to 0} [\sin (0 - h) - \cos (0 - h)]$

$= \mathop {\lim }\limits_{\scriptstyle x \to 0 - h\atop\scriptstyle h \to 0} ( - \sinh - \cosh ) = - (0) - 1 = - 1$

$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 0 - h\atop\scriptstyle h \to 0} [\sin (0 + h) - \cos (0 + h) = \mathop {\lim }\limits_{h \to 0} (\sinh - \cosh )$

lim f(x ) $=$ $\mathop {\lim }\limits_{x \to {a^ + }} [sin(0 + h)-cos(0+ h)] =$

lim(sin A-cos)
$= 0 - 1 = - 1$

Also,f(0) $=$ $- 1$
therefore, $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$

Hence, f(x) is continuous at x $=$ 0.
At $x < 0,$ f(x) $=$ sin x$-$cosx is continuous

At $x > 0,$ f(x) $=$ sin x$-$cosx is also continuous

Hence we can say that $f(x)$ is continuous at all $x \in R.$