1. **Problem statement:** We have the function $$f(x) = \frac{1}{2^x + \ln x}$$ defined on $$]0, +\infty[$$. We want to: - Prove that the 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. - Find the coordinates of the intersection point $$E$$ between the curve and the line. 2. **Asymptote at $$x=0$$:** - The line $$x=0$$ is a vertical line. - To prove it is a vertical asymptote, check the behavior of $$f(x)$$ as $$x \to 0^+$$. - Since $$\ln x \to -\infty$$ as $$x \to 0^+$$ and $$2^x \to 1$$, the denominator $$2^x + \ln x \to 1 + (-\infty) = -\infty$$. - Therefore, $$f(x) = \frac{1}{2^x + \ln x} \to 0^-$$ (approaches zero from negative side). - The function tends to zero but the denominator tends to $$-\infty$$, so the function tends to zero but the denominator tends to negative infinity. - Since the function is defined only for $$x>0$$ and tends to zero, the curve approaches the vertical line $$x=0$$ but does not cross it. - Hence, $$x=0$$ is a vertical asymptote. 3. **Limit at infinity and asymptote $$y=\frac{1}{2}x$$:** - Calculate $$\lim_{x \to +\infty} f(x)$$: - As $$x \to +\infty$$, $$2^x \to +\infty$$ much faster than $$\ln x$$. - So denominator $$2^x + \ln x \approx 2^x$$. - Thus, $$f(x) \approx \frac{1}{2^x} \to 0$$. - The limit is zero, so the function tends to zero as $$x \to +\infty$$. - Now, check if $$y=\frac{1}{2}x$$ is an asymptote: - The line $$y=\frac{1}{2}x$$ is a straight line with slope $$\frac{1}{2}$$. - To be an asymptote, the difference $$f(x) - \frac{1}{2}x$$ must tend to zero or a finite limit as $$x \to +\infty$$. - But since $$f(x) \to 0$$ and $$\frac{1}{2}x \to +\infty$$, the difference tends to $$-\infty$$. - Therefore, $$y=\frac{1}{2}x$$ is not a horizontal or oblique asymptote. - Possibly, the problem means the line $$y=0$$ is a horizontal asymptote. 4. **Intersection point $$E$$ between $$y=\frac{1}{2}x$$ and $$f(x)$$:** - Solve $$f(x) = \frac{1}{2}x$$: $$\frac{1}{2^x + \ln x} = \frac{1}{2}x$$ - Multiply both sides by $$2^x + \ln x$$: $$1 = \frac{1}{2} x (2^x + \ln x)$$ - Multiply both sides by 2: $$2 = x (2^x + \ln x)$$ - This is a transcendental equation, difficult to solve analytically. - Numerical methods or graphing are needed to find $$x$$. **Summary:** - The line $$x=0$$ is a vertical asymptote. - The limit $$\lim_{x \to +\infty} f(x) = 0$$. - The line $$y=\frac{1}{2}x$$ is not an asymptote. - The intersection point $$E$$ satisfies $$2 = x (2^x + \ln x)$$. Hence, the problem's statement about $$y=\frac{1}{2}x$$ as an asymptote seems incorrect based on the function behavior.