Let's analyze each of the given functions and determine their continuity or discontinuity:

1) $f(x) = \frac{x^2 - 16}{x - 4}$ at $x = 4$

• Simplify the expression: $f(x) = \frac{x^2 - 16}{x - 4} = \frac{(x-4)(x+4)}{x-4}$ For $x \neq 4$, the function simplifies to $f(x) = x + 4$.

• Check continuity at $x = 4$:

  • As $x \to 4$, $f(x) \to 8$ since $f(x) = x + 4$ for $x \neq 4$.
  • However, at $x = 4$, the original function $\frac{x^2 - 16}{x - 4}$ is undefined because division by zero occurs.

  Conclusion: The function has a removable discontinuity at $x = 4$. It can be redefined as $f(x) = x + 4$ for continuity at that point.

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

x^2 + 3 & \text{if } x \leq -2 \ 7 - x & \text{if } x > -2 \end{cases} ) at $x = -2$

• Check left-hand limit (as $x \to -2^-$): $\lim_{x \to -2^-} f(x) = (-2)^2 + 3 = 7$

• Check right-hand limit (as $x \to -2^+$): $\lim_{x \to -2^+} f(x) = 7 - (-2) = 9$

• Check function value: $f(-2) = (-2)^2 + 3 = 7$

  Conclusion: The left-hand limit $\lim_{x \to -2^-}$ and the right-hand limit $\lim_{x \to -2^+}$ are not equal, so the function has a jump discontinuity at $x = -2$.

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

• Check the function:

  • The function $f(x) = \frac{x + 1}{x - 2}$ is undefined at $x = 2$ since division by zero occurs.

• Check the limits:

  • As $x \to 2$, the function approaches infinity or negative infinity depending on the direction from which $x$ approaches 2.

  Conclusion: The function has an infinite discontinuity at $x = 2$.

Would you like a more detailed explanation of any of these problems? Here are some related questions you might find useful:

1. What is the difference between removable and non-removable discontinuities?
2. How can you redefine a function to remove a removable discontinuity?
3. What is the significance of checking limits from both sides of a point for continuity?
4. What are common types of discontinuities in a piecewise function?
5. How do you find the limit of a rational function as it approaches a discontinuity?

Tip: Always check both the limit and the function value at a point to confirm continuity.