If $\underset{x\to 2}{lim} \frac{\tan (x-2)\left\{x^2+(k-2)x-2k\right\}}{x^2-4x+4}=5,$ then $k$ is equal to:

$\underset{x\to 2}{lim} \frac{\tan (x-2)\left\{x^2+(k-2)x-2k\right\}}{x^2-4x+4}=5$

$\Rightarrow \underset{x\to 2}{lim} \frac{\tan (x-2)(x-2)(x+k)}{(x-2{)}^2}=5$

$\Rightarrow 1\times \underset{x\to 2}{lim} (k+x)=5$        $\left\{\because \underset{h\to 0}{lim} \frac{\tanh }{h}=1\right\}$
$$\begin{gathered} \Rightarrow k+2=5 \\ \Rightarrow k=3 \end{gathered}$$