Since the user has uploaded an image, I will first use OCR to extract the mathematical expression from the image for further processing.

import pytesseract from PIL import Image

Load the image from the file path

img_path = "/mnt/data/1725505844832mngsyfdg.jpg" img = Image.open(img_path)

Use pytesseract to extract text

extracted_text = pytesseract.image_to_string(img) extracted_textThe image text extraction didn't work correctly. However, I can directly work with the image content you've provided. The limit in question is:

$\lim_{{h \to 0}} \frac{7e^x - 7e^{x+h}}{4h}$

Let's solve this limit step by step.

Step 1: Rewrite the expression

We know that $e^{x+h} = e^x \cdot e^h$, so we can rewrite the limit as:

$\lim_{{h \to 0}} \frac{7e^x - 7e^x \cdot e^h}{4h}$

Step 2: Factor out common terms

Factor out $7e^x$ from the numerator:

$\lim_{{h \to 0}} \frac{7e^x \left(1 - e^h\right)}{4h}$

Step 3: Use the series expansion for $e^h$

For small $h$, we know the series expansion for $e^h$ is:

$e^h \approx 1 + h + \frac{h^2}{2!} + \cdots$

Substitute this into the limit expression:

$\lim_{{h \to 0}} \frac{7e^x \left(1 - \left(1 + h + \frac{h^2}{2!} + \cdots \right)\right)}{4h}$

Simplify the numerator:

$1 - \left(1 + h + \frac{h^2}{2!} + \cdots \right) = -h - \frac{h^2}{2!} - \cdots$

Thus, the expression becomes:

$\lim_{{h \to 0}} \frac{7e^x \left(-h - \frac{h^2}{2!} - \cdots \right)}{4h}$

Step 4: Simplify and evaluate the limit

Now, divide by $h$:

$\lim_{{h \to 0}} \frac{7e^x \left(-h\right)}{4h} = \lim_{{h \to 0}} \frac{-7e^x}{4}$

The higher-order terms vanish as $h \to 0$, leaving us with:

$\boxed{\frac{-7e^x}{4}}$

Thus, the value of the limit is $\frac{-7e^x}{4}$.

Let me know if you'd like any further explanation or clarification!

Here are 5 questions you might want to explore next:

1. How do higher-order terms affect limits involving exponential functions?
2. What is the general form of the derivative of $e^x$?
3. How would the solution change if the limit involved $\cos(x)$ instead of $e^x$?
4. What does the series expansion for $e^h$ tell us about small values of $h$?
5. How can L'Hopital's Rule be used to solve limits like this one?

Tip: When dealing with limits involving exponential functions, Taylor series expansions are a powerful tool for simplifying expressions around small values of the variable.