1. Consider the function $g$ defined over $]0; +\infty[$ as $g(x) = x^2 - 2\ln x$. **1) Determine $\lim_{x \to 0} g(x)$ and $\lim_{x \to +\infty} g(x)$.** - As $x \to 0^+$, $x^2 \to 0$ and $\ln x \to -\infty$, so $-2\ln x \to +\infty$. Hence, $g(x) \to +\infty$. - As $x \to +\infty$, $x^2$ dominates and $\ln x$ grows slower, so $g(x) \sim x^2 \to +\infty$. **2) Set up the table of variations and deduce $g(x) > 0$.** - Compute derivative: $g'(x) = 2x - \frac{2}{x} = \frac{2x^2 - 2}{x}$. - $g'(x)=0 \iff 2x^2 - 2 = 0 \iff x^2=1 \iff x=1$ (in domain $]0,+\infty[$). - For $x < 1$, $x^2<1 \implies g'(x)<0$, so $g$ is decreasing on $(0,1)$. - For $x > 1$, $x^2>1 \implies g'(x)>0$, so $g$ is increasing on $(1, +\infty)$. - Evaluate $g(1) = 1^2 - 2\ln 1 = 1 - 0 = 1$. - Limits: $\lim_{x \to 0^+} g(x) = +\infty$, $g(1)=1$, $\lim_{x \to +\infty} g(x) = +\infty$. - So $g$ decreases from $+\infty$ to $1$ at $x=1$, then increases back to $+\infty$. Thus $g(x) \geq 1 > 0$ for all $x > 0$. **B- Function $f$ defined by $f(x) = \frac{x}{2} + \frac{1 + \ln x}{x}$.** **1) Determine $\lim_{x \to 0} f(x)$ and deduce an asymptote to $(C)$.** - As $x \to 0^+$, $\frac{x}{2} \to 0$. - $\ln x \to -\infty$, so $1 + \ln x \to -\infty$, and dividing by $x$ (which goes to $0^+$) makes $\frac{1+\ln x}{x} \to -\infty$ (since numerator is negative and small positive denominator). - Hence, $f(x) \to -\infty$ as $x \to 0^+$. - No finite limit, so no horizontal asymptote at $0$. **2) a) Determine $\lim_{x \to +\infty} f(x)$ and show line $(\Delta): y = \frac{x}{2}$ is an asymptote.** - As $x \to +\infty$, $\frac{x}{2} \to +\infty$. - $\ln x$ grows slowly, so $\frac{1+\ln x}{x} \to 0$. - Therefore, $f(x) \sim \frac{x}{2}$ as $x \to +\infty$. - The line $y = \frac{x}{2}$ is thus an oblique asymptote. **b) Study the relative positions of $(C)$ and $(\Delta)$.** - Define $h(x) = f(x) - \frac{x}{2} = \frac{1+\ln x}{x}$. - For $x > 0$, $h(x) > 0$ if and only if $1 + \ln x > 0$. - Solve $1 + \ln x > 0 \Rightarrow \ln x > -1 \Rightarrow x > e^{-1} \approx 0.3679$. - So: - For $x > e^{-1}$, $f(x) > \frac{x}{2}$: curve above line. - For $0 < x < e^{-1}$, $f(x) < \frac{x}{2}$: curve below line. **3) Show $f'(x) = \frac{g(x)}{2x^2}$ and set up the table of variations of $f$.** - Compute $f'(x)$: $$f'(x) = \frac{1}{2} + \frac{\frac{1}{x} \cdot x - (1+\ln x) \cdot 1}{x^2} = \frac{1}{2} + \frac{1 - 1 - \ln x}{x^2} = \frac{1}{2} - \frac{\ln x}{x^2}.$$ - Rewrite $f'(x)$: $$f'(x) = \frac{x^2}{2x^2} - \frac{2 \ln x}{2 x^2} = \frac{x^2 - 2 \ln x}{2 x^2} = \frac{g(x)}{2 x^2}.$$ - Since $g(x) > 0$ for all $x$, and $x^2 > 0$, $f'(x) > 0$. - Therefore, $f$ is strictly increasing on $]0, +\infty[$. **4) Calculate coordinates of point $B$ on $(C)$ where tangent $(T)$ is parallel to $(\Delta)$.** - Tangent parallel to $(\Delta)$ means $f'(x) = \frac{1}{2}$ (slope of line $y=\frac{x}{2}$). - But $f'(x) = \frac{g(x)}{2 x^2}$. So: $$ \frac{g(x)}{2 x^2} = \frac{1}{2} \Rightarrow g(x) = x^2. $$ - Recall $g(x) = x^2 - 2 \ln x$, thus: $$ x^2 - 2 \ln x = x^2 \Rightarrow -2 \ln x = 0 \Rightarrow \ln x = 0 \Rightarrow x = 1. $$ - Compute $y$-coordinate: $$ f(1) = \frac{1}{2} + \frac{1 + \ln 1}{1} = \frac{1}{2} + \frac{1+0}{1} = \frac{1}{2} + 1 = \frac{3}{2}. $$ - So $B = (1, \frac{3}{2})$. Final answers: - $\lim_{x \to 0^+} g(x) = +\infty$, $\lim_{x \to +\infty} g(x) = +\infty$, and $g(x) > 0$. - For $f$, $\lim_{x \to 0^+} f(x) = -\infty$ (no horizontal asymptote), - Line $y=\frac{x}{2}$ is oblique asymptote as $x\to +\infty$, with curve below line for $x < e^{-1}$ and above for $x > e^{-1}$. - $f'(x) = \frac{g(x)}{2 x^2} > 0$ so $f$ strictly increasing. - Tangent parallel to asymptote at $x=1$, point $B(1, \frac{3}{2})$.