The function given is:

$f(x) = \frac{x^2 - 1}{x - 1}$

Step 1: Simplify the Function

Notice that $x^2 - 1$ is a difference of squares, which can be factored as:

$x^2 - 1 = (x - 1)(x + 1)$

So the function becomes:

$f(x) = \frac{(x - 1)(x + 1)}{x - 1}$

For $x \neq 1$, we can cancel the common term $(x - 1)$, which simplifies the function to:

$f(x) = x + 1 \quad \text{for} \quad x \neq 1$

Step 2: Analyze the Limit

Now, we need to find the limit:

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

Since the simplified function is $f(x) = x + 1$ for $x \neq 1$, we can directly substitute $x = 1$ into the simplified function:

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

Thus, the limit exists and is:

$\lim_{x \to 1} f(x) = 2$

Step 3: Behavior at $x = 1$

Even though the limit as $x \to 1$ exists, the function is undefined exactly at $x = 1$ because the original function has the term $\frac{0}{0}$ at this point, which is indeterminate. Thus, there is a hole in the graph at $x = 1$.

Step 4: Graph of the Function

• For $x \neq 1$, the function behaves like a straight line $f(x) = x + 1$.
• At $x = 1$, there is a hole in the graph.

The graph will be a straight line with a slope of 1 and a y-intercept at $(0, 1)$, but with a hole at the point $(1, 2)$.

Summary

• The limit $\lim_{x \to 1} f(x) = 2$ exists.
• The function is undefined at $x = 1$, so there is a hole in the graph at $(1, 2)$.

Would you like a detailed graph or further clarification on any part?

Here are 5 related questions:

1. What happens to the limit of a function if the denominator does not factor cleanly?
2. How do you handle limits when a function gives a $\frac{0}{0}$ indeterminate form?
3. Can limits exist if a function is undefined at a point?
4. How does factoring help in simplifying rational functions for limits?
5. What is the difference between one-sided and two-sided limits?

Tip: Always check for factoring opportunities when simplifying functions to resolve indeterminate forms.