The inequality given is:

$x^2 + 5x \leq x f(x) \leq 3x + \sin(2x)$ We are asked to find:

$\lim_{x \to 0} f(x)$

Step-by-Step Solution:

1. Consider the lower bound:

   From $x^2 + 5x \leq x f(x)$, divide both sides by $x$ (for $x \neq 0$):

   $x + 5 \leq f(x)$

   Taking the limit as $x \to 0$:

   $\lim_{x \to 0} (x + 5) = 5$

   Therefore, we have:

   $\lim_{x \to 0} f(x) \geq 5$

2. Consider the upper bound:

   From $x f(x) \leq 3x + \sin(2x)$, divide both sides by $x$ (for $x \neq 0$):

   $f(x) \leq 3 + \frac{\sin(2x)}{x}$

   We know that:

   $\lim_{x \to 0} \frac{\sin(2x)}{x} = 2$

   Therefore, as $x \to 0$:

   $f(x) \leq 3 + 2 = 5$

   Hence:

   $\lim_{x \to 0} f(x) \leq 5$

3. Conclusion:

   Since $5 \leq \lim_{x \to 0} f(x) \leq 5$, it follows that:

   $\lim_{x \to 0} f(x) = 5$

Thus, the correct answer is b) 5.

Would you like more details on any step, or do you have any other questions?

Here are five questions to expand your understanding:

1. What are the common techniques for solving limits in inequalities?
2. How does the Squeeze Theorem apply in this context?
3. Why do we divide by $x$ in the inequality, and what happens when $x = 0$?
4. How can we analyze limits of trigonometric functions like $\sin(2x)$?
5. How would the solution change if the inequality bounds were different?

Tip: Always check for continuity and use the Squeeze Theorem when faced with inequalities in limit problems.