1. **State the problem:** Prove the inequality $$0 < \frac{1}{x} \log \frac{e^x - 1}{x} < 1$$ for $$x > 0$$ using Lagrange's Mean Value Theorem (MVT). 2. **Recall Lagrange's MVT:** For a function $$f$$ continuous on $$[a,b]$$ and differentiable on $$(a,b)$$, there exists $$c \in (a,b)$$ such that $$f'(c) = \frac{f(b) - f(a)}{b - a}$$. 3. **Define the function:** Let $$g(t) = \log \frac{e^t - 1}{t}$$ for $$t > 0$$. 4. **Apply MVT on $$g$$ over $$[0,x]$$:** Since $$g$$ is continuous on $$[0,x]$$ and differentiable on $$(0,x)$$, there exists $$c \in (0,x)$$ such that $$$ g'(c) = \frac{g(x) - g(0)}{x - 0} = \frac{g(x) - g(0)}{x}. $$$ 5. **Evaluate $$g(0)$$ by limit:** $$$ g(0) = \lim_{t \to 0} \log \frac{e^t - 1}{t} = \log \lim_{t \to 0} \frac{e^t - 1}{t} = \log 1 = 0. $$$ 6. **Rewrite the expression:** $$$ \frac{1}{x} \log \frac{e^x - 1}{x} = \frac{g(x) - g(0)}{x} = g'(c). $$$ 7. **Compute $$g'(t)$$:** $$$ g(t) = \log \left( \frac{e^t - 1}{t} \right) = \log (e^t - 1) - \log t. $$$ Differentiating, $$$ g'(t) = \frac{e^t}{e^t - 1} - \frac{1}{t} = \frac{t e^t - (e^t - 1)}{t (e^t - 1)} = \frac{t e^t - e^t + 1}{t (e^t - 1)} = \frac{(t - 1) e^t + 1}{t (e^t - 1)}. $$$ 8. **Analyze the sign and bounds of $$g'(t)$$ for $$t > 0$$:** - Numerator: $$h(t) = (t - 1) e^t + 1$$. - At $$t=0$$, $$h(0) = (-1) \cdot 1 + 1 = 0$$. - Derivative $$h'(t) = e^t (t - 1) + e^t = t e^t > 0$$ for $$t > 0$$, so $$h(t)$$ is increasing and positive for $$t > 0$$. - Denominator $$t (e^t - 1) > 0$$ for $$t > 0$$. Therefore, $$g'(t) > 0$$ for $$t > 0$$. 9. **Show $$g'(t) < 1$$:** We want to prove $$$ g'(t) = \frac{(t - 1) e^t + 1}{t (e^t - 1)} < 1. $$$ Multiply both sides by denominator (positive): $$$ (t - 1) e^t + 1 < t (e^t - 1). $$$ Rewrite: $$$ (t - 1) e^t + 1 < t e^t - t. $$$ Simplify left side: $$$ (t - 1) e^t + 1 = t e^t - e^t + 1. $$$ Inequality becomes: $$$ t e^t - e^t + 1 < t e^t - t. $$$ Subtract $$t e^t$$ from both sides: $$$ - e^t + 1 < - t. $$$ Multiply both sides by $$-1$$ (reverse inequality): $$$ e^t - 1 > t. $$$ This is true for all $$t > 0$$ by the exponential function properties. 10. **Conclusion:** Since $$g'(c) = \frac{1}{x} \log \frac{e^x - 1}{x}$$ and $$0 < g'(c) < 1$$ for $$c \in (0,x)$$, the inequality $$$ 0 < \frac{1}{x} \log \frac{e^x - 1}{x} < 1 $$$ is proved. **Final answer:** $$0 < \frac{1}{x} \log \frac{e^x - 1}{x} < 1$$ for $$x > 0$$.