[tex]f(x)=\begin{cases}\frac{x}{x-2};x<2 \\ 6;2\leq x\leq6 \\ \frac{x-2}{x^2-6x+8};x>6\end{cases}[/tex]
For x = 2
[tex]\lim _{x->2^-}(\frac{x}{x-2})=-\infty[/tex][tex]\begin{gathered} \lim _{x->2^+}(6)=6 \\ \end{gathered}[/tex]
Since:
[tex]\lim _{x->2^+}f(x)\ne\lim _{x->2^-}f(x)[/tex]
The function is discontinuous at x = 2. Besides since the function has a vertical asymptote on one of the sides, we can conclude it is a infinite discontinuity.
For x = 6:
[tex]\lim _{x->6^-}6=6[/tex][tex]\lim _{x->6^+}\frac{x-2}{x^2-6x+8}=\lim _{x->6^+}\frac{x-2}{(x-2)(x-4)}=\lim _{x->6^+}\frac{1}{x-4}=\frac{1}{2}[/tex]
Since:
[tex]\lim _{x->6^+}f(x)\ne\lim _{x->6^-}f(x)[/tex]
The function is discontinuous at x = 6, at this point the function has a jump discontinuity
