We are given two limits as \( x \to 0 \) and need to find the values of \( a \) and \( b \). Let's go through each limit step by step.

### 1. First limit: \( \lim_{x \to 0} \frac{2^x - 1}{x} = a \)

This is a standard limit form. The limit \(\lim_{x \to 0} \frac{a^x - 1}{x}\) for any positive constant \( a \) is known to be \( \ln(a) \).

So, applying this result:

\[
\lim_{x \to 0} \frac{2^x - 1}{x} = \ln(2)
\]

Thus, we have:
\[
a = \ln(2)
\]

### 2. Second limit: \( \lim_{x \to 0} \frac{7^x - 1}{x} = b \)

Similarly, using the same limit form for \( 7^x \):

\[
\lim_{x \to 0} \frac{7^x - 1}{x} = \ln(7)
\]

So, we have:
\[
b = \ln(7)
\]

### Final Answers:
- \( a = \ln(2) \)
- \( b = \ln(7) \)

Would you like to see more details or have any questions?

### 5 Related Questions:
1. What is the limit of \( \lim_{x \to 0} \frac{3^x - 1}{x} \)?
2. How do we derive the limit \( \lim_{x \to 0} \frac{a^x - 1}{x} \)?
3. What is the value of \( \lim_{x \to 0} \frac{e^x - 1}{x} \)?
4. What happens to these limits if we change the base of the exponent to a fraction?
5. Can we calculate \( \lim_{x \to 1} \frac{a^x - 1}{x} \) and how does it differ?

### Tip:
The limit \( \lim_{x \to 0} \frac{a^x - 1}{x} = \ln(a) \) is related to the derivative of exponential functions, making it a key tool in calculus.

Lim x^0 = 2^x-1/x = a dan Lim x^0 = 7^x-1/x = b

Explore how to solve limit problems involving exponential functions, such as lim (2^x - 1) / x and lim (7^x - 1) / x. Understand key concepts like logarithms and the Exponential Limit Theorem, essential for students in Grades 11-12.

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