Evaluate this limit and discuss their continuity at the limiting point

When we substitute $x=1$ into the function, we get $\frac{1^2 - 1}{1 - 1} = \frac{0}{0}$, which is an indeterminate form. We can simplify the expression by factoring the numerator.
The numerator $x^2 - 1$ is a difference of squares, which factors as $(x-1)(x+1)$.

$$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x-1)(x+1)}{x - 1}$$

For $x \neq 1$, we can cancel the $(x-1)$ term:

$$\lim_{x \to 1} (x+1)$$

Now, substitute $x=1$:

$$1+1 = 2$$

So, the limit is 2.

For a function $f(x)$ to be continuous at a point $x=c$, three conditions must be met:

1. $f(c)$ must be defined.
2. $\lim_{x \to c} f(x)$ must exist.
3. $\lim_{x \to c} f(x) = f(c)$.

Let's check these conditions for $f(x) = \frac{x^2 - 1}{x - 1}$ at $x=1$:

1. Is $f(1)$ defined? Substituting $x=1$ into the original function gives $f(1) = \frac{1^2 - 1}{1 - 1} = \frac{0}{0}$, which is undefined.
2. Does $\lim_{x \to 1} f(x)$ exist? From Step 1, we found that $\lim_{x \to 1} f(x) = 2$. So, the limit exists.
3. Is $\lim_{x \to 1} f(x) = f(1)$? Since $f(1)$ is undefined, this condition cannot be satisfied.

Because $f(1)$ is undefined, the function is not continuous at $x=1$. This type of discontinuity is called a removable discontinuity because the limit exists.

The limit is $\boxed{2}$.
The function is not continuous at $x=1$ because $f(1)$ is undefined.