Here are solutions to each problem in the uploaded image:

7. Limit Problem

$\lim_{x \to 0} \frac{23e^{3x} - 23}{e^x - 1}$

Steps:

1. Factor $23$ out of the numerator: $\frac{23(e^{3x} - 1)}{e^x - 1}.$

2. Apply the first-order Taylor expansion near $x = 0$:

   • $e^{3x} \approx 1 + 3x$,
   • $e^x \approx 1 + x$.

   Substituting these approximations: $\frac{23((1 + 3x) - 1)}{(1 + x) - 1} = \frac{23(3x)}{x}.$

3. Simplify: $\frac{23(3x)}{x} = 23 \cdot 3 = 69.$

   Answer: $\boxed{69}$.

8. Trigonometric Limit

$\lim_{x \to 0} \frac{\sin(8x) \cos(4x)}{x}$

Steps:

1. Rewrite the numerator using the small-angle approximation for $\sin(kx) \approx kx$ and $\cos(kx) \approx 1$ when $x \to 0$: $\frac{\sin(8x) \cos(4x)}{x} \approx \frac{(8x)(1)}{x}.$

2. Simplify: $\frac{8x}{x} = 8.$

   Answer: $\boxed{8}$.

9. Derivative Problem

Given $f(t) = -2t^5 + \frac{3}{t^7}$, find $f'(t)$ in the form $A t^B + \frac{C}{t^D}$.

Steps:

1. Differentiate each term:

   • The derivative of $-2t^5$ is: $\frac{d}{dt}(-2t^5) = -10t^4.$

   • The derivative of $\frac{3}{t^7} = 3t^{-7}$ is: $\frac{d}{dt}(3t^{-7}) = -21t^{-8}.$

2. Combine: $f'(t) = -10t^4 - \frac{21}{t^8}.$

3. Identify coefficients and exponents:

   • $A = -10$,
   • $B = 4$,
   • $C = -21$,
   • $D = 8$.

   Answer: $A = -10$, $B = 4$, $C = -21$, $D = 8$.

Let me know if you'd like further details or have questions! Here are 5 related questions for practice:

1. How would you apply L'Hôpital's Rule to problem 7?
2. What happens if you expand higher-order terms in the Taylor series for problem 7?
3. Can you find the limit in problem 8 using L'Hôpital's Rule instead of approximations?
4. How do you verify derivatives step-by-step using the power rule?
5. How does the small-angle approximation improve limit calculations in trigonometry?

Tip:

For small $x$, always consider using Taylor expansions or small-angle approximations—they simplify calculations drastically!