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### Problem 12:
We are asked to find \(\lim_{x \to 5} f(g(x))\), given the following information:
- \(\lim_{y \to 7} f(y) = 5\),
- \(f(7) = 5\),
- \(f(5)\) is undefined,
- \(f\) is continuous,
- \(\lim_{x \to 5} g(x) = 7\).

To solve this, we use the fact that \(g(x) \to 7\) as \(x \to 5\). Since \(f\) is continuous and \(\lim_{y \to 7} f(y) = 5\), we can substitute the limit of \(g(x)\) into \(f(g(x))\):
\[
\lim_{x \to 5} f(g(x)) = f\left(\lim_{x \to 5} g(x)\right) = f(7) = 5.
\]
Thus, \(\lim_{x \to 5} f(g(x)) = 5\).

---

### Exercise 22:
The function given is:
\[
f(x) = \frac{x^2 - 9}{x^3 + 2x^2 - 9x - 18} = \frac{(x-3)(x+3)}{(x-3)(x+2)(x+3)}.
\]
Here, we need to determine the points of discontinuity.

1. **Factor the function**: As we can see, the numerator and denominator have common factors of \((x-3)\) and \((x+3)\). This simplifies the function to:
   \[
   f(x) = \frac{1}{x+2}, \quad x \neq 3, x \neq -3.
   \]
   So, the points of interest are \(x = 3\) and \(x = -3\).

2. **Check discontinuities**:
   - At \(x = 3\), there is a removable discontinuity because the factor \((x - 3)\) cancels out.
   - At \(x = -3\), the function has a vertical asymptote because the factor \((x + 3)\) does not cancel in the denominator.

Thus, \(x = 3\) is a removable discontinuity, and \(x = -3\) is a vertical asymptote (non-removable discontinuity).

---

### Exercise 23:
The piecewise function \(g(x)\) is given as:

\[
g(x) = 
\begin{cases} 
x & \text{if } x < -10, \\
0 & \text{if } x = -10, \\
\frac{x - 20}{40/(x-2)} & \text{if } -10 < x \leq 0, \\
40/(x-2) & \text{if } x > 0.
\end{cases}
\]

1. **Check for continuity**:
   - For \(x = -10\), the function switches from \(x\) to \(0\), so we need to check \(\lim_{x \to -10^-} g(x)\) and \(\lim_{x \to -10^+} g(x)\):
     - As \(x \to -10^-\), \(g(x) = x \to -10\).
     - As \(x \to -10^+\), \(\frac{x - 20}{40/(x-2)}\) needs to be checked.
     - Hence, \(g(x)\) is not continuous at \(x = -10\).

2. For \(x = 2\), \(g(x)\) is undefined since there is a division by zero in the \(x-2\) term. This point is a vertical asymptote.

---

### Problem 25:
We are asked to show that the equation \(x^6 + 5x^3 = 1\) has a solution in the interval \(I = (0,1)\).

1. Define the function:
   \[
   f(x) = x^6 + 5x^3 - 1.
   \]
   - We need to apply the Intermediate Value Theorem (IVT).
   
2. Evaluate the function at the endpoints of the interval:
   - \(f(0) = 0^6 + 5 \times 0^3 - 1 = -1\),
   - \(f(1) = 1^6 + 5 \times 1^3 - 1 = 1 + 5 - 1 = 5\).

Since \(f(0) = -1\) and \(f(1) = 5\), and \(f(x)\) is continuous, by the IVT there must be some \(c \in (0,1)\) such that \(f(c) = 0\). Thus, the equation has a solution in the interval \((0,1)\).

---

Would you like more details on any of these solutions, or help with specific concepts?

Here are 5 related questions for deeper exploration:
1. How do we know when a discontinuity is removable?
2. How can the Intermediate Value Theorem be applied to prove the existence of solutions more generally?
3. What are the conditions for a function to be continuous at a point?
4. What is the relationship between limits and continuity for piecewise functions?
5. How do vertical asymptotes affect the behavior of a function near those points?

**Tip**: When working with piecewise functions, always check for continuity and differentiability at the points where the function changes definition!

What are the limits and points of continuity for the given functions, and how can the Intermediate Value Theorem be applied to show the existence of solutions?

Explore the concept of limits and continuity, with examples involving removable discontinuities and the application of the Intermediate Value Theorem to show the existence of solutions within a given interval.

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