Prove that the function $f(x) = 5x - 3$ is continuous at x$=$ 0, at $x = 0$ and at $x = 5.$

$f(x) = 5x - 3$

At $x = 0:$

We have, $f(0) = - 3$

$\mathop{\lim }\limits_{x \to {0^ - }} f(x)$

$= \mathop {\lim }\limits_{\scriptstyle x \to 0 - h\atop\scriptstyle h \to 0} 5(0 - h) - 3$

$= - 3$

$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 0 + h\atop\scriptstyle h \to 0} 5(0 + h) - 3 = - 3$

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

Hence we can say that f is continuous at x $=$ 0.

At $x = - 3$ :
We have,$f( - 3) = 5( - 3) - 3 = - 18$

$\mathop {\lim }\limits_{x \to - {3^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 3 - h\atop\scriptstyle h \to 0} [5( - 3 - h) - 3] = \mathop {\lim }\limits_{\scriptstyle x \to 3 - h\atop\scriptstyle h \to 0} [ - 15 - 5h - 3] = - 18$

$\mathop {\lim }\limits_{x \to - {3^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 3 + h\atop\scriptstyle h \to 0} [5( - 3 + h) - 3] = \mathop {\lim }\limits_{\scriptstyle x \to 3 + h\atop\scriptstyle h \to 0} - 15 + 5h - 3] = - 18$

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

Hence we can say that f is continuous at $x = - 3.$ .

At $x = 5$ :
$f(5) = 5(5) - 3 = 22$

$\mathop {\lim }\limits_{x \to {5^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 5 - h\atop\scriptstyle h \to 0} [5(5 - h) - 3] = \mathop {\lim }\limits_{\scriptstyle x \to 5 - h\atop\scriptstyle h \to 0} 25 - 5h - 3 = 22$

$\mathop {\lim }\limits_{x \to {5^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 5 + h\atop\scriptstyle h \to 0} [5(5 + h) - 3] = \mathop {\lim }\limits_{\scriptstyle x \to 5 + h\atop\scriptstyle h \to 0} 25 + 5h - 3 = 22$

therefore, $\mathop {\lim }\limits_{x \to {5^ - }} f(x) = \mathop {\lim }\limits_{x \to {5^ + }} f(x) = f(5)$

Hence we can say that f is continuous at $x = 5$.