https://r2.erweima.ai/mathbot/image/17265755512826vfeyjxw.jpg

Let's break down and solve the limits from the image one by one.

### Problem D:
\[
D = \lim_{x \to 2} \frac{e^{2x-4} - 1}{\sqrt[3]{7x + 6} - 9}
\]
**Solution:**
1. Substitute \( x = 2 \) into the expression:
   - The numerator: \( e^{2(2) - 4} - 1 = e^0 - 1 = 0 \)
   - The denominator: \( \sqrt[3]{7(2) + 6} - 9 = \sqrt[3]{14 + 6} - 9 = \sqrt[3]{20} - 9 \)

   We get \( \frac{0}{\sqrt[3]{20} - 9} \), which is of the form \( \frac{0}{0} \), so we apply **L'Hopital's Rule**.

2. Differentiate the numerator and denominator:
   - Numerator: \( \frac{d}{dx} \left( e^{2x - 4} - 1 \right) = 2e^{2x - 4} \)
   - Denominator: \( \frac{d}{dx} \left( \sqrt[3]{7x + 6} - 9 \right) = \frac{7}{3} (7x + 6)^{-\frac{2}{3}} \)

3. Substitute \( x = 2 \) again:
   - Numerator: \( 2e^{2(2) - 4} = 2e^0 = 2 \)
   - Denominator: \( \frac{7}{3} (7(2) + 6)^{-\frac{2}{3}} = \frac{7}{3} (20)^{-\frac{2}{3}} \)

   So the limit becomes:
   \[
   \frac{2}{\frac{7}{3} (20)^{-\frac{2}{3}}} = \frac{6 (20)^{\frac{2}{3}}}{7}
   \]

### Problem E:
\[
E = \lim_{x \to 0} \frac{\sqrt{e^x + 8} - 3}{5x + 25}
\]
**Solution:**
1. Substitute \( x = 0 \):
   - Numerator: \( \sqrt{e^0 + 8} - 3 = \sqrt{9} - 3 = 3 - 3 = 0 \)
   - Denominator: \( 5(0) + 25 = 25 \)

   This results in \( \frac{0}{25} = 0 \).

Therefore, the limit is **0**.

### Problem F:
\[
F = \lim_{x \to 0} \frac{e^x + e^{2x} + e^{3x} + \dots + e^{2025x}}{2025x}
\]
**Solution:**
1. This is a sum of exponential functions. At \( x = 0 \):
   - Each term in the numerator becomes \( e^{kx} \to e^0 = 1 \) for \( k = 1, 2, 3, \dots, 2025 \).
   - The numerator has 2025 terms, each equal to 1, so the sum is \( 2025 \).

   Therefore, the limit becomes:
   \[
   \lim_{x \to 0} \frac{2025}{2025x} = \lim_{x \to 0} \frac{1}{x}
   \]
   As \( x \to 0 \), this tends to infinity, so the limit **does not exist** (or it tends to infinity).

Would you like more detailed steps on any of these, or do you have any specific questions? Here are five related questions to consider:

1. How do you apply L'Hopital's Rule when both the numerator and denominator result in \( \frac{0}{0} \)?
2. Can limits involving square roots or cube roots be solved using Taylor expansions?
3. How does the sum of exponentials behave as \( x \to 0 \) or \( x \to \infty \)?
4. What is the significance of the "form" of the limit, such as \( \frac{0}{0} \) or \( \infty/\infty \)?
5. How can the derivative of exponential functions help simplify limit calculations?

**Tip:** L'Hopital's Rule is a powerful tool for resolving indeterminate forms like \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \), but it requires differentiating both the numerator and denominator accurately.

This problem focuses on solving limits using L'Hopital's Rule and exponential functions. It explores indeterminate forms like 0/0, cube roots, and series summation of exponential terms. Includes a detailed breakdown of limits tending to infinity and zero.

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