1. **Problem statement:** We have the function $f(x) = \frac{1}{x} + \frac{\ln x}{x}$ defined for $x > 0$. We want to prove that the vertical line $x=0$ is an asymptote of the curve $C$ representing $f$, calculate $\lim_{x \to +\infty} f(x)$ and prove that the line $y=\frac{1}{2}x$ is an asymptote as $x \to +\infty$, and find the intersection point $E$ of this line with the curve. 2. **Asymptote at $x=0$:** - The vertical line $x=0$ is a vertical asymptote if $\lim_{x \to 0^+} f(x) = \pm \infty$. - Compute $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left( \frac{1}{x} + \frac{\ln x}{x} \right)$. - Since $\frac{1}{x} \to +\infty$ as $x \to 0^+$ and $\ln x \to -\infty$, but $\frac{\ln x}{x}$ behaves like $\frac{-\infty}{0^+}$ which tends to $-\infty$ slower than $\frac{1}{x}$ tends to $+\infty$, the dominant term is $\frac{1}{x}$. - Therefore, $\lim_{x \to 0^+} f(x) = +\infty$, so the line $x=0$ is a vertical asymptote. 3. **Limit at infinity and asymptote $y=\frac{1}{2}x$:** - Calculate $\lim_{x \to +\infty} \frac{f(x)}{x} = \lim_{x \to +\infty} \frac{\frac{1}{x} + \frac{\ln x}{x}}{x} = \lim_{x \to +\infty} \left( \frac{1}{x^2} + \frac{\ln x}{x^2} \right) = 0$. - Instead, consider the difference $f(x) - \frac{1}{2}x$ to check if $y=\frac{1}{2}x$ is an asymptote. - Compute $\lim_{x \to +\infty} \left( f(x) - \frac{1}{2}x \right) = \lim_{x \to +\infty} \left( \frac{1}{x} + \frac{\ln x}{x} - \frac{1}{2}x \right) = -\infty$ which is not finite. - This suggests a mistake; instead, check the limit of $\frac{f(x)}{x}$ again carefully. - Rewrite $f(x) = \frac{1 + \ln x}{x}$. - Then $\frac{f(x)}{x} = \frac{1 + \ln x}{x^2}$ which tends to 0 as $x \to +\infty$. - So $f(x)$ grows slower than any linear function. - The problem states $y=\frac{1}{2}x$ is an asymptote, so check the limit of $\frac{f(x)}{x}$ again. - Alternatively, check $\lim_{x \to +\infty} \frac{f(x)}{x} = 0$ means $f(x)$ grows slower than $x$. - The problem likely means the asymptote is $y=0$, not $y=\frac{1}{2}x$. - Re-examining the problem, the line is $y=\frac{1}{2}x$, so check $\lim_{x \to +\infty} \left( f(x) - \frac{1}{2}x \right)$. - Since $f(x)$ behaves like $\frac{\ln x}{x}$ for large $x$, which tends to 0, and $\frac{1}{2}x$ grows without bound, the difference tends to $-\infty$. - So $y=\frac{1}{2}x$ is not an asymptote. - Possibly the problem means $y=\frac{1}{2}$ (a constant), so check $\lim_{x \to +\infty} f(x)$. - Compute $\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} \frac{1 + \ln x}{x} = 0$. - So the horizontal asymptote is $y=0$. 4. **Intersection point $E$ of $y=\frac{1}{2}x$ and $f(x)$:** - Solve $f(x) = \frac{1}{2}x$. - That is $\frac{1}{x} + \frac{\ln x}{x} = \frac{1}{2}x$. - Multiply both sides by $x$: $1 + \ln x = \frac{1}{2} x^2$. - This transcendental equation can be solved numerically. - For example, try $x=1$: $1 + 0 = 1$ vs $\frac{1}{2} \times 1 = 0.5$ no. - Try $x=2$: $1 + \ln 2 \approx 1 + 0.693 = 1.693$ vs $\frac{1}{2} \times 4 = 2$ close. - Try $x=1.8$: $1 + \ln 1.8 \approx 1 + 0.588 = 1.588$ vs $\frac{1}{2} \times 3.24 = 1.62$ closer. - So $x \approx 1.8$, then $y = \frac{1}{2} \times 1.8 = 0.9$. - So $E \approx (1.8, 0.9)$. **Final answers:** - The line $x=0$ is a vertical asymptote. - The line $y=\frac{1}{2}x$ is not an asymptote; the function tends to 0 as $x \to +\infty$. - The intersection point $E$ satisfies $1 + \ln x = \frac{1}{2} x^2$, approximately $E \approx (1.8, 0.9)$. Note: The problem's statement about the asymptote $y=\frac{1}{2}x$ seems inconsistent with the function's behavior; the vertical asymptote at $x=0$ is confirmed, but the asymptote at infinity is horizontal $y=0$.