To prove the inequality $0 < \frac{1}{x} \log\left(\frac{e^x - 1}{x}\right) < 1$ for $x > 0$ using the Mean Value Theorem (MVT), we proceed as follows:

1. Define a function and apply the MVT:

Consider the function $f(x) = \log\left(\frac{e^x - 1}{x}\right)$ for $x > 0$. We want to analyze the behavior of the function $\frac{1}{x} f(x)$.

2. Prove that $f(x) > 0$ for $x > 0$:

We start by showing that $f(x) > 0$ for $x > 0$:

• When $x \to 0^+$, the limit of $\frac{e^x - 1}{x}$ is 1 (this can be shown using L'Hôpital's Rule).
• Thus, $\lim_{x \to 0^+} f(x) = \log(1) = 0$.
• For $x > 0$, $\frac{e^x - 1}{x} > 1$ (since $e^x > 1 + x$ for $x > 0$), so $f(x) = \log\left(\frac{e^x - 1}{x}\right) > 0$.

3. Use the Mean Value Theorem (MVT):

Apply the MVT to $e^t$ on the interval $[0, x]$:

• $e^x - 1 = e^c \cdot x$ for some $c \in (0, x)$.
• Thus, $\frac{e^x - 1}{x} = e^c$.

Then, the function $f(x) = \log\left(\frac{e^x - 1}{x}\right) = \log(e^c) = c$, so we can rewrite $\frac{1}{x} f(x)$ as: $\frac{1}{x} \log\left(\frac{e^x - 1}{x}\right) = \frac{c}{x}$ where $c \in (0, x)$. Since $c \in (0, x)$, we have: $0 < \frac{c}{x} < 1$ Therefore, $0 < \frac{1}{x} \log\left(\frac{e^x - 1}{x}\right) < 1$ for $x > 0$.

Conclusion:

We have shown that the function $\frac{1}{x} \log\left(\frac{e^x - 1}{x}\right)$ lies between 0 and 1 for any $x > 0$, proving the required inequality.

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. How is the Mean Value Theorem used in proving inequalities?
2. Can you prove the inequality $e^x > 1 + x$ for all $x > 0$?
3. What is the geometric interpretation of the Mean Value Theorem?
4. How can L'Hôpital's Rule be applied to evaluate limits involving logarithms?
5. Can the Mean Value Theorem be extended to functions with multiple variables?

Tip: When proving inequalities using the Mean Value Theorem, it's often useful to consider the behavior of the function at the endpoints and the derivative within the interval.