For what value of k is the following function continuous at x = 2?

Given: $f(x)=\left\{\begin{array}{c}2 x+1, \text { if } \mathrm{x}<2 \\ k, x=2 \\ 3 x-1, x>2\end{array}\right.$

We have

$(\mathrm{LHL}$ at $x=2)=\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0}(2(2-h)+1)=5$

(RHL at $x=2$ ) $=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0} 3(2+h)-1=5$

Also, $f(2)=k$

If $f(x)$ is continuous at $x=2$, then

$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2)$

$\Rightarrow 5=5=k$

Hence, for $k=5, f(x)$ is continuous at $x=2$.