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To show that each function is continuous at the specified point \( a \), we need to verify the following three conditions:

1. The function is defined at \( a \).
2. The limit of the function as \( x \) approaches \( a \) exists.
3. The limit of the function as \( x \) approaches \( a \) is equal to the function's value at \( a \).

Let's analyze each part:

### Part (a)
Given:
\[
f(x) = \begin{cases} 
\frac{x^2 - 16}{x - 4} & \text{if } x \neq 4 \\
8 & \text{if } x = 4 
\end{cases}
\quad \text{and} \quad a = 4
\]

1. **Function value at \( a = 4 \):**
   \[
   f(4) = 8
   \]

2. **Limit as \( x \) approaches 4:**

   The expression for \( f(x) \) when \( x \neq 4 \) is:
   \[
   \frac{x^2 - 16}{x - 4} = \frac{(x - 4)(x + 4)}{x - 4} = x + 4
   \]
   Therefore, 
   \[
   \lim_{x \to 4} f(x) = \lim_{x \to 4} (x + 4) = 4 + 4 = 8
   \]

3. **Check continuity:**
   Since \( \lim_{x \to 4} f(x) = f(4) = 8 \), \( f(x) \) is continuous at \( x = 4 \).

### Part (b)
Given:
\[
f(x) = \begin{cases} 
2 + 3x + 4x^2 & \text{if } x \leq 1 \\
4 + 3x + 2x^2 & \text{if } x > 1 
\end{cases}
\quad \text{and} \quad a = 1
\]

1. **Function value at \( a = 1 \):**
   \[
   f(1) = 2 + 3(1) + 4(1)^2 = 2 + 3 + 4 = 9
   \]

2. **Limit as \( x \) approaches 1 from the left (\( x \to 1^- \)):**
   \[
   \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \left(2 + 3x + 4x^2\right) = 2 + 3(1) + 4(1)^2 = 9
   \]

3. **Limit as \( x \) approaches 1 from the right (\( x \to 1^+ \)):**
   \[
   \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \left(4 + 3x + 2x^2\right) = 4 + 3(1) + 2(1)^2 = 9
   \]

4. **Check continuity:**
   Since both the left-hand limit and right-hand limit are equal, and they match the function value \( f(1) = 9 \), \( f(x) \) is continuous at \( x = 1 \).

### Part (c)
Given:
\[
f(x) = \begin{cases} 
1 - x^2 & \text{if } x \leq 1 \\
1 - x^3 & \text{if } x > 1 
\end{cases}
\quad \text{and} \quad a = 1
\]

1. **Function value at \( a = 1 \):**
   \[
   f(1) = 1 - 1^2 = 1 - 1 = 0
   \]

2. **Limit as \( x \) approaches 1 from the left (\( x \to 1^- \)):**
   \[
   \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \left(1 - x^2\right) = 1 - 1^2 = 0
   \]

3. **Limit as \( x \) approaches 1 from the right (\( x \to 1^+ \)):**
   \[
   \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \left(1 - x^3\right) = 1 - 1^3 = 0
   \]

4. **Check continuity:**
   Since both the left-hand limit and right-hand limit are equal, and they match the function value \( f(1) = 0 \), \( f(x) \) is continuous at \( x = 1 \).

### Summary
Each function \( f(x) \) is continuous at the specified point \( a \) after verifying that the function's limit as \( x \) approaches \( a \) exists and equals the function value at \( a \).

Would you like to explore any of these in more detail or have any questions?

---

Here are 5 related questions you might consider:

1. How would you determine if a function is differentiable at a point \( a \)?
2. What happens if the left-hand and right-hand limits at \( a \) are not equal?
3. Can a function be continuous at a point but not differentiable there? Provide an example.
4. How does the continuity of piecewise functions differ from standard functions?
5. What are the implications of a function being continuous on a closed interval?

**Tip:** When dealing with piecewise functions, always evaluate the limits from both sides of the point in question to ensure continuity.

Show that the following functions are continuous at the specified point: 
(a) f(x) = (x^2 - 16)/(x - 4) if x ≠ 4, 8 if x = 4, with a = 4 
(b) f(x) = 2 + 3x + 4x^2 if x ≤ 1, 4 + 3x + 2x^2 if x > 1, with a = 1 
(c) f(x) = 1 - x^2 if x ≤ 1, 1 - x^3 if x > 1, with a = 1

Explore the concept of continuity in piecewise functions through detailed examples, including verifying the continuity of functions at specific points using limits and quadratic expansion. Ideal for Grade 12 or early college students.

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