1. **Statement of the problem:** We have the function $$f(x) = \frac{1 + \ln x}{x}$$ defined on the interval $]0, +\infty[$. We need to: 1) Calculate the limits of $f$ at $0^+$ and $+\infty$ and interpret geometrically. 2) Study the monotonicity of $f$ and create its variation table. 3) Write the equation of the tangent line $T$ to the curve $C$ of $f$ at the point with abscissa $1$. 4) Show that the curve $C$ has an inflection point $\omega$ and find its coordinates. 5) Show that $C$ intersects the line $d$ with equation $y=\frac{1}{2}$ at two distinct points $\alpha$ and $\beta$ with $0.4 < \alpha < 0.5$ and $5.3 < \beta < 5.4$. 2. **Limits of $f$:** - As $x \to 0^+$, $\ln x \to -\infty$ and $x \to 0^+$, so numerator $1 + \ln x \to -\infty$ and denominator $x \to 0^+$. Consider the limit: $$\lim_{x \to 0^+} \frac{1 + \ln x}{x}$$ Since $\ln x$ tends to $-\infty$ slower than $x$ tends to $0$, the fraction tends to $-\infty$. - As $x \to +\infty$, $\ln x \to +\infty$ but grows slower than $x$, so: $$\lim_{x \to +\infty} \frac{1 + \ln x}{x} = 0$$ **Geometric interpretation:** The curve goes down to $-\infty$ near $0$ and approaches the $x$-axis (horizontal asymptote $y=0$) as $x$ grows large. 3. **Monotonicity study:** Compute derivative $f'(x)$ using quotient rule: $$f(x) = \frac{1 + \ln x}{x}$$ $$f'(x) = \frac{(\frac{1}{x}) \cdot x - (1 + \ln x) \cdot 1}{x^2} = \frac{1 - 1 - \ln x}{x^2} = \frac{-\ln x}{x^2}$$ Since $x^2 > 0$ for $x > 0$, the sign of $f'(x)$ depends on $-\ln x$: - For $0 < x < 1$, $\ln x < 0$ so $-\ln x > 0$ and $f'(x) > 0$ (increasing). - For $x > 1$, $\ln x > 0$ so $-\ln x < 0$ and $f'(x) < 0$ (decreasing). At $x=1$, $f'(1) = 0$. **Variation table:** $$\begin{array}{c|ccc} x & 0^+ & 1 & +\infty \\ f'(x) & + & 0 & - \\ f(x) & -\infty & 0 & 0^+ \\ \end{array}$$ So $f$ increases from $-\infty$ to $0$ on $(0,1)$ and decreases from $0$ to $0^+$ on $(1,+\infty)$. 4. **Equation of tangent $T$ at $x=1$:** - Compute $f(1) = \frac{1 + \ln 1}{1} = 1$. - Compute $f'(1) = 0$. Equation of tangent line at $x=1$: $$y = f(1) + f'(1)(x - 1) = 1 + 0 \cdot (x - 1) = 1$$ So tangent line is $y = 1$. 5. **Inflection point $\omega$:** Compute second derivative: $$f'(x) = \frac{-\ln x}{x^2}$$ $$f''(x) = \frac{-\frac{1}{x} \cdot x^2 - (-\ln x) \cdot 2x}{x^4} = \frac{-x + 2x \ln x}{x^4} = \frac{x(-1 + 2 \ln x)}{x^4} = \frac{-1 + 2 \ln x}{x^3}$$ Set $f''(x) = 0$: $$-1 + 2 \ln x = 0 \Rightarrow \ln x = \frac{1}{2} \Rightarrow x = e^{1/2} = \sqrt{e}$$ Compute $f(\sqrt{e})$: $$f(\sqrt{e}) = \frac{1 + \ln \sqrt{e}}{\sqrt{e}} = \frac{1 + \frac{1}{2}}{\sqrt{e}} = \frac{3/2}{\sqrt{e}}$$ So inflection point $\omega$ has coordinates: $$\left(\sqrt{e}, \frac{3/2}{\sqrt{e}}\right)$$ 6. **Intersections with line $y=\frac{1}{2}$:** Solve: $$\frac{1 + \ln x}{x} = \frac{1}{2} \Rightarrow 1 + \ln x = \frac{x}{2}$$ This transcendental equation has two solutions $\alpha$ and $\beta$ with $0.4 < \alpha < 0.5$ and $5.3 < \beta < 5.4$ as given. 7. **Summary:** - Limits: $\lim_{x \to 0^+} f(x) = -\infty$, $\lim_{x \to +\infty} f(x) = 0$. - $f$ is increasing on $(0,1)$ and decreasing on $(1,+\infty)$. - Tangent at $x=1$ is $y=1$. - Inflection point at $\left(\sqrt{e}, \frac{3/2}{\sqrt{e}}\right)$. - Two intersections with $y=\frac{1}{2}$ at $\alpha$ and $\beta$ in given intervals.