We are given the inequality:

$x^2 + 4x - 1 \leq f(x) \leq x^2 - 4x + 7 \quad \text{for all} \, x \, \text{near} \, x = 2, \text{except possibly at} \, x = 2.$

We are asked to evaluate the limit of $f(x)$ as $x \to 2$.

Step 1: Evaluate the two bounding functions at $x = 2$

First, we calculate the values of the two bounding functions at $x = 2$:

• For the left-hand side function $x^2 + 4x - 1$: $(2)^2 + 4(2) - 1 = 4 + 8 - 1 = 11.$

• For the right-hand side function $x^2 - 4x + 7$: $(2)^2 - 4(2) + 7 = 4 - 8 + 7 = 3.$

So, we now have:

$3 \leq f(x) \leq 11 \quad \text{for all} \, x \, \text{near} \, 2.$

Step 2: Apply the Squeeze Theorem

The Squeeze Theorem tells us that if $f(x)$ is squeezed between two functions that both approach the same limit as $x \to 2$, then $\lim_{x \to 2} f(x)$ must also approach that same limit.

• The left-hand side function $x^2 + 4x - 1$ approaches 11 as $x \to 2$.
• The right-hand side function $x^2 - 4x + 7$ approaches 3 as $x \to 2$.

Since the two bounding functions do not approach the same value at $x = 2$, the Squeeze Theorem does not apply, and we cannot conclude that $f(x)$ will approach a single value.

Thus, the limit does not exist.

Final Answer:

The correct answer is:

C.) limit as $x \to 2$ of $f(x)$ does not exist.

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If you'd like further clarification, feel free to ask!

Here are some related questions to expand on the topic:

1. What is the Squeeze Theorem and how is it used in limit evaluation?
2. How do we evaluate the limit of a function when it's bounded by two other functions?
3. What happens if two bounding functions approach different limits at the same point?
4. What are other methods to evaluate limits of functions when direct substitution doesn’t work?
5. How do we apply the definition of a limit to confirm whether a limit exists?

Tip: Always check if the two bounding functions approach the same limit before applying the Squeeze Theorem!