from PIL import Image import pytesseract

Load the image

image_path = '/mnt/data/1732300870682mvvisgyh.jpg' image = Image.open(image_path)

Use OCR to extract the mathematical expression

extracted_text = pytesseract.image_to_string(image) extracted_text'lum) Sc\n=\n\nyork\n''lum) Sc\n=\n\nyork\n'successThe OCR process did not correctly interpret the mathematical expression due to the quality or format of the image. Let me interpret it manually:

The problem appears to be: $\lim_{x \to -\infty} \sqrt{x^2 + 2x + x - 1}$

I will solve it.To solve:

$\lim_{x \to -\infty} \sqrt{x^2 + 2x + x - 1}$

Step 1: Simplify the expression inside the square root

Combine like terms inside the square root: $x^2 + 2x + x - 1 = x^2 + 3x - 1$

So the expression becomes: $\lim_{x \to -\infty} \sqrt{x^2 + 3x - 1}$

Step 2: Factor out $x^2$ from the terms inside the square root

For large $x$, factor $x^2$ from the terms: $\sqrt{x^2 + 3x - 1} = \sqrt{x^2 \left(1 + \frac{3}{x} - \frac{1}{x^2}\right)}$

Step 3: Simplify the square root

Using the property $\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$, we simplify: $\sqrt{x^2 \left(1 + \frac{3}{x} - \frac{1}{x^2}\right)} = |x| \cdot \sqrt{1 + \frac{3}{x} - \frac{1}{x^2}}$

Since $x \to -\infty$, $|x| = -x$, so the expression becomes: $-x \cdot \sqrt{1 + \frac{3}{x} - \frac{1}{x^2}}$

Step 4: Evaluate the limit

As $x \to -\infty$:

• $\frac{3}{x} \to 0$
• $\frac{1}{x^2} \to 0$

Thus, the term inside the square root approaches: $\sqrt{1 + 0 - 0} = \sqrt{1} = 1$

So the expression becomes: $-x \cdot 1 = -x$

Finally, as $x \to -\infty$, $-x \to +\infty$.

Final Answer:

$\lim_{x \to -\infty} \sqrt{x^2 + 2x + x - 1} = +\infty$

Would you like further clarification? Here are some related questions for exploration:

1. What happens to this limit as $x \to +\infty$?
2. How can we confirm the result graphically?
3. Why does the factorization of $x^2$ work in simplifying limits?
4. Can this method be applied to any polynomial under a square root?
5. What if the expression had higher-order terms in the numerator?

Tip: Always factor out the highest power of $x$ when handling square root limits as $x$ approaches infinity.