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The problem presented is the following limit:

\[
\lim_{x \to 0} \frac{e^{-ax} - e^{-bx}}{x^2}
\]

To solve this, we can apply a Taylor series expansion around \( x = 0 \) for each exponential term in the numerator, since this will help simplify the expression as \( x \to 0 \).

### Step-by-Step Solution

1. **Expand \( e^{-ax} \) and \( e^{-bx} \) using the Taylor series around \( x = 0 \):**
   \[
   e^{-ax} \approx 1 - ax + \frac{(ax)^2}{2} + O(x^3)
   \]
   \[
   e^{-bx} \approx 1 - bx + \frac{(bx)^2}{2} + O(x^3)
   \]

2. **Subtract the two expansions:**
   \[
   e^{-ax} - e^{-bx} \approx \left(1 - ax + \frac{(ax)^2}{2}\right) - \left(1 - bx + \frac{(bx)^2}{2}\right)
   \]
   Simplifying each term:
   \[
   e^{-ax} - e^{-bx} \approx -(a - b)x + \frac{1}{2}((a^2 - b^2)x^2) + O(x^3)
   \]

3. **Rewrite the expression inside the limit:**
   Substitute the approximation into the original limit:
   \[
   \lim_{x \to 0} \frac{-(a - b)x + \frac{1}{2}(a^2 - b^2)x^2}{x^2}
   \]

4. **Separate terms in the numerator:**
   \[
   = \lim_{x \to 0} \left(\frac{-(a - b)x}{x^2} + \frac{\frac{1}{2}(a^2 - b^2)x^2}{x^2}\right)
   \]

   Simplifying each term:
   \[
   = \lim_{x \to 0} \left(-\frac{a - b}{x} + \frac{1}{2}(a^2 - b^2)\right)
   \]

5. **Take the limit as \( x \to 0 \):**
   As \( x \to 0 \), the term \( -\frac{a - b}{x} \) diverges unless \( a = b \). Therefore, for the limit to exist finitely, we must have \( a = b \).

   If \( a = b \), the expression becomes:
   \[
   \lim_{x \to 0} \frac{e^{-ax} - e^{-ax}}{x^2} = \lim_{x \to 0} \frac{0}{x^2} = 0
   \]

### Conclusion
The limit is \( 0 \) if \( a = b \). If \( a \neq b \), the limit does not exist (it diverges).

---

Would you like further details on any part of this solution? Here are some related questions to deepen your understanding:

1. What is the Taylor series expansion for an exponential function?
2. Why does the limit diverge when \( a \neq b \)?
3. How can we use Taylor expansions to approximate limits?
4. What happens if we have higher powers of \( x \) in the denominator?
5. How does L'Hôpital's Rule apply to limits involving exponential terms?

**Tip:** When facing limits with exponential functions, Taylor series expansion is a useful tool for simplification around points where the function behavior is tricky, like \( x = 0 \).

Find the limit as x approaches 0 of (e^(-ax) - e^(-bx)) / x^2.

Learn how to evaluate the limit of (e^(-ax) - e^(-bx)) / x^2 as x approaches 0 using Taylor series expansion. This problem covers key calculus concepts and techniques, ideal for students in Grades 11-12.

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