Evaluate this limit and discuss their continuity at the limiting point

To evaluate the limit and discuss continuity, we will first simplify the function and then apply the conditions for continuity.

Step 1: Evaluate the limit $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$. 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 $2$.

Step 2: Discuss the continuity of the function at $x=1$. 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.

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