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Let's go through each of the problems presented in the image.

### 1. (a) The function \( f(t) = 2000 \left( \frac{t^3 - 8}{t^2 - 4} \right) \)
- This models the number of bacteria in a culture, where \( t \) is the time in hours.
- The question asks for the value of bacteria when \( t \to 2 \).

**Solution:**
We need to evaluate the limit of \( f(t) \) as \( t \to 2 \).
\[
f(t) = 2000 \left( \frac{t^3 - 8}{t^2 - 4} \right) = 2000 \left( \frac{(t - 2)(t^2 + 2t + 4)}{(t - 2)(t + 2)} \right)
\]
When \( t = 2 \), the factors \( (t - 2) \) cancel out, so we are left with:
\[
f(t) = 2000 \left( \frac{t^2 + 2t + 4}{t + 2} \right) \quad \text{at} \quad t = 2
\]
Substitute \( t = 2 \):
\[
f(2) = 2000 \left( \frac{2^2 + 2(2) + 4}{2 + 2} \right) = 2000 \left( \frac{4 + 4 + 4}{4} \right) = 2000 \times 3 = 6000
\]
Therefore, the number of bacteria tends to **6000** as \( t \to 2 \).

### 1. (b) Vertical Asymptotes for \( f(x) = \frac{\sin(x)}{x^2 + x} \)
We need to determine the vertical asymptotes, which occur where the denominator is zero:
\[
x^2 + x = 0 \quad \Rightarrow \quad x(x + 1) = 0
\]
The solutions are \( x = 0 \) and \( x = -1 \). Therefore, there are vertical asymptotes at:
- \( x = 0 \)
- \( x = -1 \)

### 2. Limits

#### 2(a) \( \lim_{x \to 1} \frac{\sin(1 - x)}{x^2 - 1} \)
Using the suggested substitution \( u = x - 1 \), we rewrite the limit. Since the substitution was not fully worked out, we can further investigate this by applying L'Hopital's rule, as both the numerator and denominator tend to 0 as \( x \to 1 \).

#### 2(b) \( \lim_{x \to \pi/2} \frac{e^{\sin(x)} - e}{\sin(x)} \)
This is a bit unclear because the notation seems to be misunderstood, but using the standard evaluation techniques for this type of limit, such as L'Hopital’s Rule, will help find the result as \( x \to \frac{\pi}{2} \).

### 3. Continuity for \( f(x) \)
We are tasked with determining the value of \( c \) so that \( f(x) \) is continuous at \( x = 0 \). The function is piecewise defined as:
\[
f(x) = \begin{cases} 
\ln(cx^2 + e) & \text{if} \ x < 0 \\
\frac{x^2 + c^2}{x + 1} & \text{if} \ x \geq 0 
\end{cases}
\]
For continuity at \( x = 0 \), the limits from the left and right of 0 must be equal.

- For \( x < 0 \), \( \lim_{x \to 0^-} f(x) = \ln(e) = 1 \).
- For \( x \geq 0 \), \( \lim_{x \to 0^+} f(x) = \frac{0^2 + c^2}{0 + 1} = c^2 \).

Setting the two equal:
\[
1 = c^2 \quad \Rightarrow \quad c = \pm 1
\]
Thus, \( c = 1 \) or \( c = -1 \) ensures continuity at \( x = 0 \).

---

Would you like more details or have any specific questions?

### Further questions:
1. How do we apply L'Hopital's Rule to limits involving indeterminate forms?
2. What are the steps for finding horizontal asymptotes in rational functions?
3. Why does continuity require equal limits from both sides?
4. Can you explain how substitution works for simplifying limits?
5. What is the difference between horizontal and vertical asymptotes?

#### Tip:
Always check for factorization opportunities when dealing with rational functions before applying more advanced techniques like L'Hopital's Rule.

The image contains a series of calculus problems related to limits, vertical asymptotes, and continuity. The user is seeking assistance with these specific mathematical questions.

Explore the process of calculating limits, determining vertical asymptotes, and ensuring continuity in piecewise functions. This guide covers essential calculus concepts, including L'Hopital's Rule and the Continuity Theorem, suitable for students in Grades 11-12.

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