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The limit you provided is:

\[
\lim_{x \to 0} \frac{5^x - 7^x}{x}
\]

### Steps to solve:

1. **Recognize Indeterminate Form**: When we plug in \( x = 0 \), both \( 5^x \) and \( 7^x \) approach 1, so the numerator becomes \( 5^0 - 7^0 = 1 - 1 = 0 \). Thus, the expression is of the form \( \frac{0}{0} \), which is indeterminate. We can apply **L'Hopital's Rule**.

2. **L'Hopital's Rule**: This rule states that for indeterminate forms \( \frac{0}{0} \), we can compute the limit by differentiating the numerator and denominator with respect to \( x \).

   - Differentiate the numerator: \( \frac{d}{dx}(5^x - 7^x) = 5^x \ln 5 - 7^x \ln 7 \).
   - Differentiate the denominator: \( \frac{d}{dx}(x) = 1 \).

   Applying L'Hopital's Rule, the limit becomes:
   \[
   \lim_{x \to 0} \frac{5^x \ln 5 - 7^x \ln 7}{1} = 5^0 \ln 5 - 7^0 \ln 7 = \ln 5 - \ln 7.
   \]

3. **Final Answer**:
   \[
   \lim_{x \to 0} \frac{5^x - 7^x}{x} = \ln 5 - \ln 7.
   \]

Do you want further details or have any questions? Here are some related questions:

1. How does L'Hopital's Rule apply to different indeterminate forms?
2. What are the other techniques for solving limits besides L'Hopital’s Rule?
3. Can you explain why \( 5^x \) and \( 7^x \) approach 1 as \( x \to 0 \)?
4. What are the conditions under which L'Hopital's Rule can be used?
5. How does the natural logarithm \( \ln \) appear in derivatives of exponential functions?

**Tip**: When solving limits that result in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), always consider L'Hopital's Rule as a potential approach!

Calculate the limit of (5^x - 7^x)/x as x approaches 0 by applying L'Hopital's Rule. This problem involves understanding indeterminate forms and the derivative of exponential functions. Suitable for students learning undergraduate calculus.

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