The image contains mathematical exercises in French focusing on calculus and analysis. I'll help you by solving one or more parts of the exercises systematically and providing detailed explanations for each step. Let's begin!

Exercise I:

1. a) Montrer que : g'(x) = (x + 1)/x pour tout x de I.

The function given is:

$g(x) = x - 1 + \ln x$

To compute $g'(x)$, differentiate each term:

• Derivative of $x$: $1$
• Derivative of $-1$: $0$
• Derivative of $\ln x$: $1/x$

Thus: $g'(x) = 1 + \frac{1}{x} = \frac{x + 1}{x}$

Conclusion: $g'(x) = \frac{x + 1}{x}$, which holds for all $x \in I$.

1. b) Montrer que la fonction $g$ est croissante sur $I$.

A function is increasing on an interval if its derivative is positive. From the previous result: $g'(x) = \frac{x + 1}{x}$

For $x > 0$:

• $x + 1 > 0$ and $x > 0$, so $g'(x) > 0$.

Conclusion: $g(x)$ is strictly increasing on $I = ]0; +\infty[$.

2. En déduire que $g(x) \geq 0$ sur $[1; +\infty[$ et $g(x) \leq 0$ sur $]0; 1]$.

• Since $g$ is increasing on $]0; +\infty[$, its value at $x = 1$ gives the boundary condition. Compute $g(1)$: $g(1) = 1 - 1 + \ln(1) = 0$

• For $x > 1$, $g(x) > g(1) = 0$, so $g(x) \geq 0$ on $[1; +\infty[$.

• For $0 < x < 1$, $g(x) < g(1) = 0$, so $g(x) \leq 0$ on $]0; 1]$.

Exercise II:

1. a) Montrer que $\lim_{x \to 0^+} f(x) = +\infty$ et donner une interprétation géométrique.

The function is: $f(x) = \left( \frac{x - 1}{x} \right) \ln x$

Rewrite $f(x)$: $f(x) = \left( 1 - \frac{1}{x} \right) \ln x$

Analyze $\ln x$ and $1/x$ as $x \to 0^+$:

• $\ln x \to -\infty$
• $-\frac{1}{x} \to -\infty$

Thus, $f(x) \to +\infty$ because $\ln x$ dominates near 0.

Interpretation: As $x \to 0^+$, the curve $C$ rises steeply toward $+\infty$.

1. b) Montrer que $\lim_{x \to +\infty} f(x) = +\infty$ et $\lim_{x \to +\infty} \frac{f(x)}{x} = 0$.

To compute $\lim_{x \to +\infty} f(x)$:

• As $x \to +\infty$, $\ln x \to +\infty$.
• Since $\frac{x - 1}{x} = 1 - \frac{1}{x} \to 1$, $f(x) \to +\infty$.

To compute $\lim_{x \to +\infty} \frac{f(x)}{x}$: $\frac{f(x)}{x} = \frac{(x - 1)\ln x}{x^2} = \frac{\ln x}{x} - \frac{\ln x}{x^2}$

• Both terms $\frac{\ln x}{x}$ and $\frac{\ln x}{x^2} \to 0$ as $x \to +\infty$.

Thus: $\lim_{x \to +\infty} \frac{f(x)}{x} = 0$

Let me know if you would like me to solve part II.2 or provide more details!

Relative Questions:

1. Why is $g'(x) > 0$ sufficient to prove $g(x)$ is increasing?
2. Can $g(x)$ be negative for $x > 1$? Why or why not?
3. What does the interpretation of $\lim_{x \to 0^+} f(x) = +\infty$ mean geometrically?
4. How does $\ln x$'s growth compare to polynomials as $x \to +\infty$?
5. Why does $\frac{\ln x}{x^2} \to 0$ faster than $\frac{\ln x}{x}$?

Tip: For limits involving $\ln x$,