1. **Problem statement:** We have the function $f(x) = 2x(1 - \ln x)$ defined on $]0, +\infty[$. We want to find the limits of its derivative $f'(x)$ as $x \to 0$ and $x \to +\infty$. 2. **Derivative calculation:** Using the product rule, $f'(x) = \frac{d}{dx}[2x(1 - \ln x)] = 2(1 - \ln x) + 2x \cdot \frac{d}{dx}(1 - \ln x)$. 3. Since $\frac{d}{dx}(1 - \ln x) = -\frac{1}{x}$, we get $$f'(x) = 2(1 - \ln x) + 2x \left(-\frac{1}{x}\right) = 2(1 - \ln x) - 2 = 2 - 2\ln x - 2 = -2 \ln x.$$ 4. **Limits of $f'(x)$:** - As $x \to 0^+$, $\ln x \to -\infty$, so $f'(x) = -2 \ln x \to +\infty$. - As $x \to +\infty$, $\ln x \to +\infty$, so $f'(x) = -2 \ln x \to -\infty$. 5. **Finding point A (intersection with x-axis):** Solve $f(x) = 0$: $$2x(1 - \ln x) = 0 \implies x=0 \text{ (not in domain) or } 1 - \ln x = 0 \implies \ln x = 1 \implies x = e.$$ So, $A = (e, 0)$. 6. **Confirming $f'(x) = -2 \ln x$:** Already shown in step 3. 7. **Table of variations:** Since $f'(x) = -2 \ln x$: - For $x \in ]0,1[$, $\ln x < 0$ so $f'(x) > 0$ (increasing). - At $x=1$, $f'(1) = -2 \cdot 0 = 0$ (critical point). - For $x > 1$, $\ln x > 0$ so $f'(x) < 0$ (decreasing). Calculate $f(1) = 2 \cdot 1 (1 - 0) = 2$. Thus, $f$ increases on $(0,1)$ from $0$ to $2$, then decreases on $(1, +\infty)$. 8. **Equation of tangent (T) at A:** - Slope at $A$ is $f'(e) = -2 \ln e = -2 \cdot 1 = -2$. - Point $A$ is $(e,0)$. - Equation: $y - 0 = -2(x - e)$ or $$y = -2x + 2e.$$