Classify the discontinuity at x = 1

Factor the denominator: $x^2-1 = (x-1)(x+1)$, so for $x\neq 1$, $c(x) = \frac{x-1}{(x-1)(x+1)} = \frac{1}{x+1}.$ This simplification is valid everywhere except at $x=1$, where the original expression is $0/0$.

Taking the limit using the simplified form, $\lim_{x\to 1} c(x) = \lim_{x\to 1} \frac{1}{x+1} = \frac{1}{2}$, so the limit exists. However, $c(1)$ itself is undefined since the original function has $0$ in the denominator there.

Since the limit exists but $f(1)$ is not defined, the condition 'f(a) must be defined' fails while the limit condition holds. This means the discontinuity at $x=1$ is removable — the graph has a hole at $(1, 1/2)$ that could be patched by defining $c(1) = 1/2$.

At $x=-1$, the factor $(x+1)$ in the denominator does not cancel with anything in the numerator $x-1$, since $x-1$ does not equal zero at $x=-1$. So unlike at $x=1$, there is no common factor to remove.

As $x \to -1$, the denominator $x+1 \to 0$ while the numerator approaches $-2$, so the one-sided limits grow without bound (one side toward $+\infty$ and the other toward $-\infty$), meaning the limit does not exist in the finite sense.

Because at least one one-sided limit is infinite, this corresponds to a vertical asymptote at $x=-1$, which is an infinite discontinuity — a fundamentally different type from the removable hole found at $x=1$.