The problem is asking to prove a necessary and sufficient condition for the limit of a strictly increasing function $f(x)$ on $(0, +\infty)$ to approach $A$ as $x \to +\infty$, given that $\lim_{n \to +\infty} f(x_n) = A$.

Step-by-step solution:

1. Given assumptions:

   • $f(x)$ is strictly increasing on $(0, +\infty)$.
   • $\lim_{n \to +\infty} f(x_n) = A$, where $A$ is a finite constant.

2. Goal:

   • Prove that the necessary and sufficient condition for $\lim_{x \to +\infty} f(x) = A$ is that $\lim_{n \to +\infty} x_n = +\infty$.

Proof:

(1) Sufficiency:

• If $\lim_{x \to +\infty} f(x) = A$, then for any sequence $\{x_n\} \to +\infty$, we also have $f(x_n) \to A$ because $f(x)$ is strictly increasing and continuous in its limit behavior.

(2) Necessity:

• Assume $\lim_{n \to +\infty} f(x_n) = A$. To ensure $\lim_{x \to +\infty} f(x) = A$, we must prove $\lim_{n \to +\infty} x_n = +\infty$.
• Since $f(x)$ is strictly increasing, it is invertible on $(0, +\infty)$. Let $f^{-1}$ be the inverse of $f$.
• By the property of limits: x_n = f^{-1}(f(x_n)) \to f^{-1}(A), \quad \text{if A is finite.}
• For $f(x) \to A$ as $x \to +\infty$, $f^{-1} \to +\infty$ as $A \to$ the upper bound of $f(x)$. Hence, $x_n \to +\infty$.

Thus, the condition $\lim_{n \to +\infty} x_n = +\infty$ is necessary and sufficient for $\lim_{x \to +\infty} f(x) = A$.

Would you like further clarification or additional examples?
Here are five related questions to deepen your understanding:

1. What properties of strictly increasing functions are crucial in proving this result?
2. How would the proof change if $f(x)$ were not strictly increasing but monotonic?
3. What happens if $A$ is not finite? Can the proof still hold?
4. How does the invertibility of $f(x)$ contribute to the proof?
5. Can you think of a counterexample where $\lim_{n \to +\infty} f(x_n) = A$, but $\lim_{x \to +\infty} f(x) \neq A$?

Tip: Always check the behavior of $f(x)$ at infinity by analyzing its growth (e.g., linear, logarithmic, exponential).