. $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{ - 2,if}&{x \le - 1}\\{2x,if}&{ - 1 < x \le 1}\\{2,if}&{x > 1}\end{array}} \right.$

At $x = - 1:$

R.H.L.$= \mathop {\lim }\limits_{x \to - {1^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to - 1 + h\atop\scriptstyle h \to 0} 2( - 1 + h) = - 2$

L.H.L. $= \mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to - 1 - h\atop\scriptstyle h \to 0} ( - 2) = - 2$

Also, $f( - 1) = - 2$
therefore, $\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} f(x) = f( - 1)$

Hence, $f(x)$ is continuous at x$=$ $- 1$ .
At x $=$ 1:

R.H.L.$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} (2) = 2$

L.H.L$= \mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 1 - h\atop\scriptstyle h \to 0} 2(1 - h) = 2$

Also, f(1) $=$ 2

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

Hence we can say that f(x) is continuous at x $=$ 1.