Answer
For $a=2$ the limit exists and is finite.
Work Step by Step
Firstly, simplify the limit $\lim_\limits{x\to 1}\left(\dfrac{1}{x-1}-\dfrac{a}{x^2-1}\right)$ as follows:
$\lim_\limits{x\to 1}\left(\dfrac{1}{x-1}-\dfrac{a}{x^2-1}\right)=\lim_\limits{x\to 1}\left(\dfrac{x+1-a}{x^2-1}\right)=\lim_\limits{x\to 1}\left(\dfrac{x+1-a}{(x-1)(x+1)}\right)$
Since, if $x \to1$, then $x-1$ approaches $0$. Which is in the denominator.
So, the limit will exists only if $x-1$ is a factor of the numerator.
That is, $x+1-a=x-1$ or $a=2$
Substitute $a=2$ in the limit and solve
$\lim_\limits{x\to 1}\left(\dfrac{x+1-2}{(x-1)(x+1)}\right)=\lim_\limits{x\to 1}\left(\dfrac{x-1}{(x-1)(x+1)}\right)=\lim_\limits{x\to 1}\left(\dfrac{1}{x+1}\right)=\dfrac{1}{2}$
Hence, for $a=2$ the limit exists and is finite.
