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### Question 4(a):
Given 
\[ f(x) = \begin{cases} 
\frac{x^3 - 64}{x - 4}, & x \neq 4, \\
40, & x = 4.
\end{cases} \]

#### (i) Find \(\lim_{x \to 4} f(x)\).

We notice that for \(x \neq 4\), the function \(f(x)\) is a rational expression:
\[
\frac{x^3 - 64}{x - 4}
\]
The numerator \(x^3 - 64\) is a difference of cubes, which can be factored as:
\[
x^3 - 64 = (x - 4)(x^2 + 4x + 16)
\]
So, the expression for \(f(x)\) simplifies to:
\[
f(x) = \frac{(x - 4)(x^2 + 4x + 16)}{x - 4} = x^2 + 4x + 16 \quad \text{for} \, x \neq 4.
\]
Thus, for \(x \neq 4\), we have:
\[
f(x) = x^2 + 4x + 16
\]
Now, to find \(\lim_{x \to 4} f(x)\), substitute \(x = 4\) into the simplified expression:
\[
f(4) = 4^2 + 4(4) + 16 = 16 + 16 + 16 = 48
\]
So:
\[
\lim_{x \to 4} f(x) = 48
\]

#### (ii) Is \(f(x)\) continuous at \(x = 4\)? Give your reason.

To check if \(f(x)\) is continuous at \(x = 4\), we need to verify:
1. \(f(4)\) exists.
2. \(\lim_{x \to 4} f(x)\) exists.
3. \(f(4) = \lim_{x \to 4} f(x)\).

From the definition of \(f(x)\), we know:
\[
f(4) = 40
\]
From part (i), we found that:
\[
\lim_{x \to 4} f(x) = 48
\]
Since \(f(4) \neq \lim_{x \to 4} f(x)\), \(f(x)\) is **not continuous** at \(x = 4\).

---

### Question 4(b):
Given the piecewise function:
\[
h(x) = \begin{cases} 
Ax - B, & x \leq -1, \\
2x^2 + 3Ax + B, & -1 < x \leq 1, \\
4, & x > 1,
\end{cases}
\]
Determine the values of \(A\) and \(B\) such that \(h(x)\) is continuous for all \(x\).

For \(h(x)\) to be continuous, the following conditions must hold:
1. \(\lim_{x \to -1^-} h(x) = \lim_{x \to -1^+} h(x)\)
2. \(\lim_{x \to 1^-} h(x) = \lim_{x \to 1^+} h(x)\)

#### Step 1: Continuity at \(x = -1\)
From the left side (\(x \leq -1\)), we use the expression \(Ax - B\). So:
\[
\lim_{x \to -1^-} h(x) = A(-1) - B = -A - B
\]
From the right side (\(-1 < x \leq 1\)), we use the expression \(2x^2 + 3Ax + B\). Substituting \(x = -1\):
\[
\lim_{x \to -1^+} h(x) = 2(-1)^2 + 3A(-1) + B = 2 - 3A + B
\]
For continuity at \(x = -1\), we set these limits equal:
\[
-A - B = 2 - 3A + B
\]
Simplifying:
\[
-2A - 2B = 2 \quad \Rightarrow \quad A + B = -1 \tag{1}
\]

#### Step 2: Continuity at \(x = 1\)
From the left side (\(-1 < x \leq 1\)), we again use \(2x^2 + 3Ax + B\). Substituting \(x = 1\):
\

Given f(x) with different values for x = 4 and x ≠ 4, find the limit of f(x) as x approaches 4 and determine if the function is continuous at x = 4. Also, determine values of A and B for the continuity of the function h(x) defined in different pieces.

Explore how to solve limits and check the continuity of functions. This example covers finding the limit of a piecewise function at specific points and determining continuity by solving for constants A and B.

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