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!