I see you've uploaded an image containing mathematical questions. I’ll analyze the image and solve the questions step by step. Let’s begin. PIL import Image import pytesseract

Load the image from the user-uploaded file

image_path = '/mnt/data/17339193673098clvlkhp.jpg' image = Image.open(image_path)

Use OCR to extract text from the image

extracted_text = pytesseract.image_to_string(image, lang='fra')

extracted_textIt seems there was an issue with processing the text due to a language setting problem. Let me retry using English as the default OCR language to extract the mathematical content. Retry OCR using the default English language setting extracted_text_en = pytesseract.image_to_string(image, lang='eng')

extracted_text_enThe OCR extraction wasn't perfect, but I can still work with the visible structure and questions to interpret the mathematics manually. Let me clean and interpret the content for clarity.

From the questions visible:

1. Q2 asks for the limit $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2 + x^3}$.
2. Q3-Q6 involve expansions (développement limité) or equivalent forms at specific points.

I'll solve these systematically. Starting with Q2.### Solving Q2: We are tasked to compute the limit: $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2 + x^3}.$

Step 1: Expand $e^x$ using its Taylor series around $x = 0$:

$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \mathcal{O}(x^4).$

Thus: $e^x - 1 - x = \frac{x^2}{2} + \frac{x^3}{6} + \mathcal{O}(x^4).$

Step 2: Substitute into the numerator:

$\frac{e^x - 1 - x}{x^2 + x^3} = \frac{\frac{x^2}{2} + \frac{x^3}{6} + \mathcal{O}(x^4)}{x^2 + x^3}.$

Step 3: Factor out $x^2$ in the denominator:

$\frac{e^x - 1 - x}{x^2 + x^3} = \frac{\frac{x^2}{2} + \frac{x^3}{6} + \mathcal{O}(x^4)}{x^2(1 + x)}.$

Step 4: Simplify:

Divide numerator and denominator by $x^2$: $\frac{e^x - 1 - x}{x^2 + x^3} = \frac{\frac{1}{2} + \frac{x}{6} + \mathcal{O}(x^2)}{1 + x}.$

Expand $\frac{1}{1+x}$ as $1 - x + \mathcal{O}(x^2)$: $\frac{e^x - 1 - x}{x^2 + x^3} = \left(\frac{1}{2} + \frac{x}{6} + \mathcal{O}(x^2)\right)(1 - x + \mathcal{O}(x^2)).$

Step 5: Combine terms:

$\frac{e^x - 1 - x}{x^2 + x^3} = \frac{1}{2} - \frac{x}{2} + \frac{x}{6} + \mathcal{O}(x^2).$

As $x \to 0$, only the constant term remains: $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2 + x^3} = \frac{1}{2}.$

Final Answer:

The correct choice is D. $\frac{1}{2}$.

Would you like me to continue with Q3–Q6? Let me know if you'd like detailed steps or a summary for each!