The given image presents a table and a limit problem. It seems the objective is to complete the table and estimate the limit of the function:

$\lim_{x \to 4} \frac{x - 4}{x^2 - 5x + 4}$

This limit can be rewritten as:

$\lim_{x \to 4} \frac{x - 4}{(x - 4)(x - 1)} = \lim_{x \to 4} \frac{1}{x - 1}$

Now, let's compute the values at the different $x$ points provided in the table using this simplified form.

Expression for $f(x)$:

$f(x) = \frac{1}{x - 1}$

Using this, we can complete the table by plugging in the values of $x$:

1. $x = 3.9$: $f(3.9) = \frac{1}{3.9 - 1} = \frac{1}{2.9} \approx 0.3448$
2. $x = 3.99$: $f(3.99) = \frac{1}{3.99 - 1} = \frac{1}{2.99} \approx 0.3345$
3. $x = 3.999$: $f(3.999) = \frac{1}{3.999 - 1} = \frac{1}{2.999} \approx 0.3334$
4. $x = 4$: $f(4)$ is undefined, because $f(x)$ involves division by zero when $x = 4$.
5. $x = 4.001$: $f(4.001) = \frac{1}{4.001 - 1} = \frac{1}{3.001} \approx 0.3334$
6. $x = 4.01$: $f(4.01) = \frac{1}{4.01 - 1} = \frac{1}{3.01} \approx 0.3345$
7. $x = 4.1$: $f(4.1) = \frac{1}{4.1 - 1} = \frac{1}{3.1} \approx 0.3226$

Conclusion:

The limit as $x \to 4$ for this function is: $\lim_{x \to 4} \frac{1}{x - 1} = \frac{1}{3} \approx 0.3333$

Thus, the correct answer is $0.3333$, not $\frac{1}{3}$, and the table has been mostly filled correctly except for the special case when $x = 4$ where the function is undefined.

Would you like further details or have any specific questions on this?

Here are 5 relative questions that might help you understand the concept better:

1. Why is the function undefined at $x = 4$?
2. What happens if you try to directly substitute $x = 4$ into the original function?
3. How does the table approach help in estimating the limit of a function?
4. What are some common types of functions that result in limits involving indeterminate forms?
5. Why is the limit result $\frac{1}{3}$ important, and how does it relate to the behavior of the function near $x = 4$?

Tip: Always remember to check if the function simplifies before computing limits, especially when encountering indeterminate forms like $\frac{0}{0}$.