For [tex]f[/tex] to be continuous at [tex]x=5[/tex], we need to have
[tex]\displaystyle\lim_{x\to5}f(x)=f(5)[/tex]
Note that [tex]x\to5[/tex] means that [tex]x\neq5[/tex], but that [tex]x[/tex] is *approaching* 5. We're told that for [tex]x\neq5[/tex], we have
[tex]f(x)=\dfrac{x^2-5x}{x^2-25}[/tex]
We can write
[tex]\dfrac{x^2-5x}{x^2-25}=\dfrac{x(x-5)}{(x+5)(x-5)}=\dfrac x{x+5}[/tex]
and the limit would be
[tex]\displaystyle\lim_{x\to5}\frac x{x+5}=\dfrac5{5+5}=\dfrac5{10}=\dfrac12\neq1=f(5)[/tex]
and so [tex]f[/tex] is discontinuous.
