To solve the problem using the Squeeze Theorem, let’s proceed step by step.

Problem Setup

The function $f(x)$ is bounded as follows for $x \in \mathcal{D}(f)$ (domain of $f$):

$\frac{\sin\left( -\frac{\pi}{20}x \right)}{x^2 + 10x + 25} \leq f(x) \leq \frac{\sqrt{2} \cdot 2^{x+5}}{2x^2 + 20x + 50}.$

We are tasked to compute the following limit using the Squeeze Theorem:

$\lim_{x \to -5} (x+5)^2 f(x).$

Step 1: Analyze the Bounds

1. First, simplify the denominator in the lower bound:

   • $x^2 + 10x + 25 = (x+5)^2$.
   • Therefore, the lower bound becomes: $\frac{\sin\left( -\frac{\pi}{20}x \right)}{(x+5)^2}.$

2. Simplify the denominator in the upper bound:

   • $2x^2 + 20x + 50 = 2(x^2 + 10x + 25) = 2(x+5)^2$.
   • Therefore, the upper bound becomes: $\frac{\sqrt{2} \cdot 2^{x+5}}{2(x+5)^2}.$

Thus, the inequality for $f(x)$ can be rewritten as:

$\frac{\sin\left( -\frac{\pi}{20}x \right)}{(x+5)^2} \leq f(x) \leq \frac{\sqrt{2} \cdot 2^{x+5}}{2(x+5)^2}.$

Step 2: Multiply by $(x+5)^2$

To isolate $f(x)$, multiply through the inequality by $(x+5)^2$ (note: $(x+5)^2 > 0$ near $x = -5$):

$\sin\left( -\frac{\pi}{20}x \right) \leq (x+5)^2 f(x) \leq \frac{\sqrt{2} \cdot 2^{x+5}}{2}.$

Step 3: Take the Limit as $x \to -5$

1. Behavior of the lower bound:

   • As $x \to -5$, $\sin\left( -\frac{\pi}{20}x \right)$ oscillates between $-1$ and $1$.
   • Thus, the lower bound approaches a value between $-1$ and $1$.

2. Behavior of the upper bound:

   • $2^{x+5}$ becomes $2^0 = 1$ as $x \to -5$.
   • Therefore, the upper bound becomes: $\frac{\sqrt{2} \cdot 1}{2} = \frac{\sqrt{2}}{2}.$

Thus, near $x = -5$, we have:

$-1 \leq \lim_{x \to -5} (x+5)^2 f(x) \leq \frac{\sqrt{2}}{2}.$

Step 4: Apply the Squeeze Theorem

The Squeeze Theorem states that if a function is squeezed between two values that both approach the same limit, then the function itself approaches that limit.

Here, as $x \to -5$, the lower bound oscillates between $-1$ and $1$, and the upper bound approaches $\frac{\sqrt{2}}{2}$. Since the lower bound remains finite and the upper bound approaches a small positive value, the limit of $(x+5)^2 f(x)$ must be 0.

Final Answer

$\lim_{x \to -5} (x+5)^2 f(x) = 0.$

Do you want more details on any step, or need a further explanation? Let me know!

Here are 5 related questions:

1. What is the Squeeze Theorem, and how is it applied?
2. Why does the limit of $\sin\left( -\frac{\pi}{20}x \right)$ oscillate between $-1$ and $1$?
3. How do you simplify quadratic expressions like $x^2 + 10x + 25$?
4. What happens when you multiply inequalities by positive versus negative values?
5. Why does the term $2^{x+5}$ simplify to $1$ as $x \to -5$?

Tip: Always check for factorizations of quadratic expressions to simplify limits involving denominators.