The mathematical problem involves applying the Squeeze Theorem (Prinsip Apit) to find the limit of an expression containing the floor function $\lfloor 2x \rfloor$.

Here’s a step-by-step breakdown:

1. Understanding the inequalities:

We are given: $x - 1 < \lfloor x \rfloor \leq x$ This is the basic property of the floor function, which states that the greatest integer less than or equal to $x$ is $\lfloor x \rfloor$, and it lies between $x-1$ and $x$.

2. Applying it to $2x$:

Multiplying the above inequality by 2, we get: $2x - 1 < \lfloor 2x \rfloor \leq 2x$

3. Using the Squeeze Theorem (Prinsip Apit):

We aim to calculate: $\lim_{x \to \infty} \frac{\lfloor 2x \rfloor}{5x + 2}$ Now, using the inequality $2x - 1 < \lfloor 2x \rfloor \leq 2x$, divide the entire inequality by $5x + 2$: $\frac{2x - 1}{5x + 2} < \frac{\lfloor 2x \rfloor}{5x + 2} \leq \frac{2x}{5x + 2}$

4. Taking the limit:

Now, take the limit as $x \to \infty$ for both the left and right side of the inequality.

• For the left side: $\lim_{x \to \infty} \frac{2x - 1}{5x + 2} = \lim_{x \to \infty} \frac{2 - \frac{1}{x}}{5 + \frac{2}{x}} = \frac{2}{5}$

• For the right side: $\lim_{x \to \infty} \frac{2x}{5x + 2} = \lim_{x \to \infty} \frac{2}{5 + \frac{2}{x}} = \frac{2}{5}$

Since the left-hand side and right-hand side both approach $\frac{2}{5}$, by the Squeeze Theorem, we conclude: $\lim_{x \to \infty} \frac{\lfloor 2x \rfloor}{5x + 2} = \frac{2}{5}$

Conclusion:

Thus, the limit is: $\lim_{x \to \infty} \frac{\lfloor 2x \rfloor}{5x + 2} = \frac{2}{5}$

Would you like more details or have any questions about the steps? Here are some follow-up questions that can deepen your understanding:

1. How does the Squeeze Theorem (Prinsip Apit) work in general for limits?
2. What are the properties of the floor function $\lfloor x \rfloor$ that were used here?
3. Why does multiplying the inequalities by 2 preserve the relationship?
4. What would happen if we used different functions in place of the floor function in this limit problem?
5. How does adding constants in the denominator affect the result of the limit?

Tip: When dealing with floor functions in limits, it's important to first bound the function using its properties and then apply the Squeeze Theorem to simplify complex limits.