Discuss the continuity of the following functions :

(a) $f(x) = \sin x + \cos x$

(b) $f(x) = \sin x - \cos x$

(c) $f(x) = \sin x \cdot \cos x$

(a) Let a be an arbitrary real number. Then,
$f(a) = \sin a + \cos a$

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

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

$= \sin a(1) + \cos a(0) + \cos a(1) - \sin a(0) = \sin a + \cos a$

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

$= \mathop {\lim }\limits_{\scriptstyle x \to a - h\atop\scriptstyle h \to 0} [(\sin a\cosh - \cos a\sinh ) + (\cos a\cosh + \sin a\sinh )]$
$= \sin a(1) - \cos a(0) + \cos a(1) + \sin a(0) = \sin a + \cos a.$

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

$\Rightarrow$ $f(x)$ is continuous at $x = a$.

Hence we can say that $f(x) =$ sin x + cos x is everywhere continuous.

(b) Let a be an arbitrary real number. Then$f(a) = \sin a - \cos a$

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

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

$= \sin a(1) + \cos a(0) - \cos a(1) + \sin a(0) = \sin a - \cos a$

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

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

$= \sin a(1) - \cos a(0) - \cos a(1) - \sin a(0)$
$= \sin a - \cos a.$

therefore, $\mathop {\lim }\limits_{x \to {a^ - }} f(x) = f(a) = \mathop {\lim }\limits_{x \to {a^ = }} f(x)$

$\Rightarrow$ f(x) is continuous at $x = a$ .

Hence we can say that $f(x) = \sin x - \cos x$ is everywhere continuous.

(c) Let a be an arbitrary real number. Then, $f(a) = \sin a\cos$a

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

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

$= ((\sin a(1) + \cos a(0)((\cos a)(1) - \sin a(0))$
$= \sin a\cos a.$

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

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

$= (\sin (a)(1) - \cos a(0))(\cos a(1) + \sin a(0)))$
$= \sin a\cos a$

therefore,$\mathop {\lim }\limits_{x \to {a^ - }} f(x) = f(a) = \mathop {\lim }\limits_{x \to {a^ + }} f(x).$

$\Rightarrow$ $f(x)$ is continuous at $x = a$.

Hence we can say that f(x)$= \sin x \cdot \cos x$ is everywhere continuous.