Identifying a removable discontinuity via limits

For $x \neq 1$, factor the numerator: $x^2 - 1 = (x-1)(x+1)$, so $\frac{x^2-1}{x-1} = x+1$ for all $x \neq 1$. Therefore $\lim_{x\to 1} q(x) = \lim_{x\to 1}(x+1) = 2$.

Since $\lim_{x\to 1} q(x) = 2$ exists but $q(1) = 3$, the limit does not equal the function value, so $q$ is discontinuous at $x=1$. Because the limit exists (finite) and differs only from the assigned function value, this is a removable discontinuity: redefining $q(1) = 2$ would make $q$ continuous there.