To evaluate the limit

$\lim_{x \to 1} \frac{f(x) - 3}{x - 1}$

we need to understand what this expression represents. This is a form of the derivative definition, specifically evaluating $f'(1)$, the derivative of $f(x)$ at $x = 1$. In other words, we are looking for the slope of the tangent line to the graph of $f(x)$ at $x = 1$.

From the graph, we can determine the following key points near $x = 1$:

1. $f(1) = 3$, as indicated in the graph (the function value at $x = 1$ is 3).
2. To find the slope around $x = 1$, observe the change in $f(x)$ as $x$ increases from 1.

By estimating the slope between points near $x = 1$:

• At $x = 0$, $f(x) = 2$.
• At $x = 2$, $f(x) = 4$.

So the slope can be approximated by looking at the change in $f(x)$ over this interval: $\text{slope} = \frac{f(2) - f(0)}{2 - 0} = \frac{4 - 2}{2 - 0} = 1$

Thus, $\lim_{x \to 1} \frac{f(x) - 3}{x - 1} = 1$

Therefore, the answer is:

$\boxed{1}$

Would you like a further breakdown of the steps involved, or do you have any specific questions?

Here are some related questions to deepen understanding:

1. How would the limit change if the slope of the graph at $x = 1$ were different?
2. What does this limit tell us about the behavior of $f(x)$ around $x = 1$?
3. Can we apply this method of finding the derivative to other points on the graph?
4. How does this approach relate to the formal definition of a derivative?
5. What if $f(x)$ had a discontinuity at $x = 1$? How would the problem change?

Tip: When evaluating limits that resemble the form of a derivative, recognize that the expression might represent the slope of a tangent line at a point, allowing you to use graphical or derivative methods.