We are given two functions $g(x) = -2x^2 + 4x + 1$ and $h(x) = \frac{1}{2}x^2 - x + \frac{11}{2}$. The problem asks for $\lim_{x \to 1} f(x)$, where $f(x)$ is a function that satisfies:

$g(x) \leq f(x) \leq h(x) \quad \text{for all } x.$

This situation hints at applying the Squeeze Theorem (also known as the Sandwich Theorem) because $f(x)$ is "squeezed" between two functions, $g(x)$ and $h(x)$. To apply the Squeeze Theorem, we need to evaluate the limits of $g(x)$ and $h(x)$ as $x \to 1$ and see if they are equal.

Step 1: Evaluate $\lim_{x \to 1} g(x)$

$g(x) = -2x^2 + 4x + 1$ Substituting $x = 1$: $g(1) = -2(1)^2 + 4(1) + 1 = -2 + 4 + 1 = 3$ Thus, $\lim_{x \to 1} g(x) = 3$.

Step 2: Evaluate $\lim_{x \to 1} h(x)$

$h(x) = \frac{1}{2}x^2 - x + \frac{11}{2}$ Substituting $x = 1$: $h(1) = \frac{1}{2}(1)^2 - 1 + \frac{11}{2} = \frac{1}{2} - 1 + \frac{11}{2} = \frac{1}{2} + \frac{11}{2} - 1 = 6 - 1 = 5$ Thus, $\lim_{x \to 1} h(x) = 3$.

Step 3: Apply the Squeeze Theorem

Since $g(x) \leq f(x) \leq h(x)$ and both $\lim_{x \to 1} g(x) = 3$ and $\lim_{x \to 1} h(x) = 3$, by the Squeeze Theorem, we can conclude: $\lim_{x \to 1} f(x) = 3$

Thus, the limit is $\boxed{3}$.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. What is the general Squeeze Theorem, and when can it be applied?
2. How can we visually interpret the squeezing of a function between two bounds on a graph?
3. Could the limit change if $g(x)$ and $h(x)$ had different limits as $x \to 1$?
4. How does the concept of continuity tie into the Squeeze Theorem?
5. What are the necessary conditions for the Squeeze Theorem to work?

Tip: The Squeeze Theorem is especially useful when you cannot directly compute the limit of a function but can bound it between two other functions whose limits are easier to find.