Is the function defined by $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{x + 5,}&{ifx \le 1}\\{x - 5,}&{ifx > 1}\end{array}} \right.$ a continuous function?}}

We observe thatf(x) is continuous at all real numbers x < 1 and x > 1 as it is polynomial function.
Now, continuity at x $=$ 1:

$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} (x - 5) = \mathop {\lim }\limits_{\scriptstyle x \to 1 + h\atop\scriptstyle h \to 0} (1 + h - 5) = \mathop {\lim }\limits_{\scriptstyle x \to 1 + h\atop\scriptstyle h \to 0} (h - 4) = - 4$

$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} (x + 5) = \mathop {\lim }\limits_{\scriptstyle x \to 1 - h\atop\scriptstyle h \to 0} (1 - h + 5) = \mathop {\lim }\limits_{\scriptstyle x \to 1 - h\atop\scriptstyle h \to 0} (6 - h) = 6$

Thus, $\mathop {\lim }\limits_{x \to {1^ + }} f(x) \ne \mathop {\lim }\limits_{x \to {1^ - }} f(x)$

Hence we can say that f(x) is not continuous at x $=$ 1.