$f(x) = \left\{ {\begin{array}{cccccccccccccccccccc}{3x - 8,{\rm{ if }}x \le 5}\\{2k,{\rm{ }}\,\,\,\,\,\,\,{\rm{if }}x > 5}\end{array}} \right.$at $x = 5$ }

We have $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{3x - 8,}&{{\rm{ if }}x \le 5}\\{2k,}&{{\rm{ if }}x > 5}\end{array}} \right.$at $x = 5$

Since, $f(x)$ is continuous at $x = 5$.

therefore,${\rm{LHL}} = {\rm{RHL}} = f(5)$

Now, ${\rm{LHL}} = \mathop {\lim }\limits_{x \to {5^ - }} (3x - 8) = \mathop {\lim }\limits_{h \to 0} [3(5 - h) - 8]$

$= \mathop {\lim }\limits_{h \to 0} [15 - 3h - 8] = 7$

${\rm{RHL}} = \mathop {\lim }\limits_{x \to {5^ + }} 2{\rm{k}} = \mathop {\lim }\limits_{h \to 0} 2{\rm{k}} = 2{\rm{k}} = 7$
and $f(5) = 3 \times 5 - 8 = 7$

$\Rightarrow 2k = 7 \Rightarrow k = \frac{7}{2}$

Therefore the value of k =$\frac{7}{2}$