[tex]f(x)=\dfrac{x+1}{x^2-1}[/tex] has two points of discontinuity at [tex]x=\pm1[/tex], where [tex]x^2-1=0[/tex].
We have
[tex]\dfrac{x+1}{x^2-1}=\dfrac{x+1}{(x+1)(x-1)}[/tex]
so as long as [tex]x\neq-1[/tex], we can cancel [tex]x+1[/tex] and be left with [tex]\dfrac1{x-1}[/tex]. Then as [tex]x\to-1[/tex], we have
[tex]\displaystyle\lim_{x\to-1}f(x)=\lim_{x\to-1}\frac1{x-1}=-\frac12[/tex]
The limit exists, so the discontinuity at -1 is removable (so definitely not C or D).
Meanwhile,
[tex]\displaystyle\lim_{x\to1}f(x)=\lim_{x\to1}\frac1{x-1}[/tex]
does not exist, so the discontinuity at 1 is non-removable, making A the correct answer.
