The problem asks to determine if each function is continuous at the given $x$-values. If not continuous, the discontinuity should be classified.

Let's analyze each function one by one.

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

The function $f(x) = \frac{x^2 - 1}{x - 1}$ can be simplified:

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

For $x \neq 1$, $f(x) = x + 1$. At $x = 1$, the function is undefined because the denominator becomes zero, which suggests a removable discontinuity at $x = 1$. If the function is redefined as $f(1) = 2$, it would be continuous.

1b) $f(x) = x^2 - 2x + 3$ at $x = 1$

This is a polynomial function, which is continuous everywhere. Therefore, $f(x)$ is continuous at $x = 1$.

2a) $f(x) = \frac{2x + 3}{x - 2}$ at $x = 2$

The function $f(x) = \frac{2x + 3}{x - 2}$ is undefined at $x = 2$ because the denominator becomes zero. This suggests an infinite discontinuity at $x = 2$, where the function tends to infinity as $x$ approaches 2.

2b) ( f(x) = \begin{cases}

2x + 1 & \text{if } x < 1 \ x^2 & \text{if } x \geq 1 \end{cases} ) at $x = 1$

We need to check the left-hand limit, right-hand limit, and the value of the function at $x = 1$:

• $\lim\limits_{x \to 1^-} f(x) = 2(1) + 1 = 3$
• $\lim\limits_{x \to 1^+} f(x) = 1^2 = 1$
• $f(1) = 1^2 = 1$

Since the left-hand limit does not equal the right-hand limit, $f(x)$ has a jump discontinuity at $x = 1$.

3a) $f(x) = \frac{x^2 - x - 6}{x + 2}$ at $x = -2$

The function $f(x) = \frac{x^2 - x - 6}{x + 2}$ can be simplified by factoring the numerator:

$f(x) = \frac{(x - 3)(x + 2)}{x + 2}$

For $x \neq -2$, $f(x) = x - 3$. At $x = -2$, the function is undefined because the denominator becomes zero, indicating a removable discontinuity at $x = -2$.

3b) $f(x) = \frac{x + 1}{x^2 - 4}$ at $x = -2$

The function $f(x) = \frac{x + 1}{x^2 - 4}$ can be factored as:

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

At $x = -2$, the denominator becomes zero, so the function is undefined, which indicates an infinite discontinuity at $x = -2$.

Would you like more detailed explanations, or do you have any other questions?

Here are 5 questions to consider:

1. How do you determine if a function has a removable discontinuity?
2. What is the difference between a jump discontinuity and an infinite discontinuity?
3. How can you redefine a function to make it continuous at a point with a removable discontinuity?
4. Why are polynomial functions always continuous?
5. How do you approach analyzing piecewise functions for continuity?

Tip: When dealing with rational functions, always check if the function can be simplified to identify potential removable discontinuities.