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.