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 $x = 0$.