$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{kx + 1,}&{if}&{x \le 5}\\{3x - 5,}&{if}&{x > 5}\end{array}} \right.$ at $x = 5$

$\mathop {\lim }\limits_{x \to {5^ - }} f(x) = \mathop {\lim }\limits_{x \to {5^ - }} (kx + 1)$

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

$\mathop {\lim }\limits_{x \to {5^ + }} f(x) = \mathop {\lim }\limits_{x \to {5^ + }} (3x - 5) = \mathop {\lim }\limits_{\scriptstyle x \to 5 + h\atop\scriptstyle h \to 0} (3(5 + h) - 5) = 3(5 + 0) - 5 = 10$

For continuity at x$= 5,\mathop {\lim }\limits_{x \to {5^ - }} f(x) = \mathop {\lim }\limits_{x \to {5^ + }} f(x) = f(5)$
$\Rightarrow$ $5k + 1 = 10 \Rightarrow 5k = 9 \Rightarrow k = \cfrac{9}{5}$