Examine the continuity of the function f(x)= $2{x^2} - 1$ at $x = 3.$

$f(x) = 2{x^2} - 1;R.H.L.$

$= \mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 3 + h\atop\scriptstyle h \to 0} 2{(3 + h)^2} - 1$

$= \mathop {\lim }\limits_{\scriptstyle x \to 3 + h\atop\scriptstyle h \to 0} 2(9 + 6h + {h^2}) - 1$

$= \mathop {\lim }\limits_{\scriptstyle x \to 3 + h\atop\scriptstyle h \to 0} (18 + 12h + 2{h^2}) - 1 = \mathop {\lim }\limits_{\scriptstyle x \to 3 + h\atop\scriptstyle h \to 0} (17 + 12h + 2{h^2}) = 17$

$L.H.L. = \mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 3 - h\atop\scriptstyle h \to 0} 2{(3 - h)^2} - 1$

$= \mathop {\lim }\limits_{\scriptstyle x \to 3 - h\atop\scriptstyle h \to 0} 2(9 - 6h + {h^2}) - 1$

$= \mathop {\lim }\limits_{\scriptstyle x \to 3 - h\atop\scriptstyle h \to 0} (18 - 12h + 2{h^2}) - 1$

$= \mathop {\lim }\limits_{\scriptstyle x \to 3 - h\atop\scriptstyle h \to 0} 2{h^2} - 12h + 17 = 17$

therefore, $R.H.L. = L.H.L.$

Also, $f(3) = 2{(3)^2} - 1 = 17$

therefore, $\mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{x \to {3^ - }} f(x) = f(3)$

Hence we can say that the given function $f(x) = 2{x^2} - 1$ is continuous at x $=$ 3.