The image you've uploaded shows a problem involving limits. The specific problem asks for a comment about the value of the limit as $x$ approaches 3 of the function given by $\frac{x^4 - 1}{x^3 - x}$, and the mathematical work involves the use of L'Hôpital's Rule to evaluate the limit.

Let's analyze and solve the limit step by step:

Problem

$\lim_{x \to 3} \frac{x^4 - 1}{x^3 - x}$

Step 1: Direct Substitution

First, substitute $x = 3$ into the function to check if it yields an indeterminate form.

$\text{Numerator: } 3^4 - 1 = 81 - 1 = 80$ $\text{Denominator: } 3^3 - 3 = 27 - 3 = 24$

Since the numerator and denominator do not produce an indeterminate form (0/0 or ∞/∞), direct substitution can be used to find the limit:

$\lim_{x \to 3} \frac{x^4 - 1}{x^3 - x} = \frac{80}{24} = \frac{10}{3}$

Thus, the limit is $\frac{10}{3}$.

Comment on the Calculation

The calculation initially presented seems to be an application of L'Hôpital's Rule, which is used for indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. However, in this case, direct substitution was sufficient because the limit does not produce an indeterminate form.

Additional Questions and Tip

Would you like more details on the solution? If so, here are some questions that might help you delve deeper:

1. What are the conditions under which L'Hôpital's Rule can be applied?
2. Can you identify other methods to solve this limit apart from direct substitution?
3. How would the solution change if the limit approached a different value, such as 0?
4. What does the limit represent in a real-world context?
5. How can we factor the numerator and denominator to simplify the expression before substitution?

Tip: When working with limits, always check for the possibility of directly substituting the value first. L'Hôpital's Rule should be applied only when an indeterminate form is encountered.