To evaluate the limit $\lim_{x \to 2} \frac{f(x) - 3}{x - 2}$ using L'Hôpital's Rule, we proceed as follows:

Given:

1. $f(2) = 3$
2. The slope of the graph of $f(x)$ at $x = 2$ is $\frac{1}{2}$. This slope represents the derivative $f'(2) = \frac{1}{2}$.

The limit is in the form $\frac{0}{0}$ because:

• As $x \to 2$, $f(x) \to f(2) = 3$, so the numerator $f(x) - 3 \to 0$.
• The denominator $x - 2 \to 0$.

Since the limit is indeterminate, we can apply L'Hôpital's Rule, which states that if $\lim_{x \to a} \frac{g(x)}{h(x)} = \frac{0}{0}$, then:

$\lim_{x \to a} \frac{g(x)}{h(x)} = \lim_{x \to a} \frac{g'(x)}{h'(x)}$

For our limit, we let $g(x) = f(x) - 3$ and $h(x) = x - 2$.

Step 1: Differentiate the Numerator and Denominator

1. The derivative of the numerator $g(x) = f(x) - 3$ is $g'(x) = f'(x)$.
2. The derivative of the denominator $h(x) = x - 2$ is $h'(x) = 1$.

Thus, applying L'Hôpital's Rule, we get:

$\lim_{x \to 2} \frac{f(x) - 3}{x - 2} = \lim_{x \to 2} \frac{f'(x)}{1} = f'(2)$

Step 2: Substitute the Known Value of $f'(2)$

We are given that $f'(2) = \frac{1}{2}$.

So,

$\lim_{x \to 2} \frac{f(x) - 3}{x - 2} = \frac{1}{2}$

Final Answer

$\lim_{x \to 2} \frac{f(x) - 3}{x - 2} = \frac{1}{2}$

Would you like further details on this solution or have additional questions?

---

Here are five related questions to explore further:

1. How does L'Hôpital's Rule apply in cases of other indeterminate forms like $\infty / \infty$?
2. What if the limit provided different values for $f(2)$ or $f'(2)$; how would that impact the result?
3. Can you find limits that do not require L'Hôpital's Rule even if they initially look indeterminate?
4. How can we interpret the result of this limit in terms of the behavior of the function near $x = 2$?
5. How does the derivative represent the slope of the tangent line at a point, and why is it important in limits?

Tip: Always check if a limit can be simplified before applying L'Hôpital's Rule; sometimes algebraic simplifications make it unnecessary.