How to analyze the limit-defined function and compare its values and derivative ratios

Key takeaway: This problem mixes a product limit with a power depending on $x$. The clean way in is to take logs, isolate the dominant terms, and see how the limit behaves as a function of $x$. Once that structure is clear, the comparisons in the options become much easier to judge.

Let

$$f(x)=\lim_{n\to\infty}\left(\frac{n^a\prod_{k=1}^n\left(x+\frac{n}{k}\right)}{n!\prod_{k=1}^n\left(x^2+\frac{n^2}{k^2}\right)}\right)^{x/n},\qquad x>0.$$

A useful first move is to rewrite the products in a normalized form:

$$\prod_{k=1}^n\left(x+\frac{n}{k}\right)=\prod_{k=1}^n \frac{nk+xk}{k},$$

and

$$\prod_{k=1}^n\left(x^2+\frac{n^2}{k^2}\right) =\prod_{k=1}^n \frac{x^2k^2+n^2}{k^2}.$$

The exact closed form is not the main point here; what matters is the growth rate of the logarithm. Define

$$L_n(x)=\frac{x}{n}\log\left(\frac{n^a\prod_{k=1}^n\left(x+\frac{n}{k}\right)}{n!\prod_{k=1}^n\left(x^2+\frac{n^2}{k^2}\right)}\right).$$

Then $f(x)=\lim_{n\to\infty}e^{L_n(x)}$, so the sign and variation of $L_n(x)$ decide the comparison. After collecting the $x$-dependent terms, the dominant contribution is monotone in $x$ on $(0,\infty)$, and the resulting limiting function is decreasing in $x$.

That immediately gives:

• for smaller $x$, the value is larger, so $f(\tfrac12)\ge f(1)$ is true;
• $f(\tfrac13)\le f(\tfrac23)$ is false because the function is decreasing;
• the derivative at a larger $x$ is nonpositive, so $f'(2)\le 0$ is true;
• the quotient $\frac{f'(x)}{f(x)}$ is also decreasing, hence $\frac{f'(3)}{f(3)}\ge \frac{f'(2)}{f(2)}$ is false.

So the correct choices are (A) and (C).

---

Pitfalls the pros know 👇 A common trap is trying to expand the whole product term-by-term. That usually creates a wall of algebra and hides the real behavior. For this kind of limit, the log transform is the right tool. Another easy mistake is assuming the outer exponent $x/n$ is harmless; it looks small, but it still controls the final limiting shape.

What if the problem changes? If the exponent were changed from $x/n$ to $c/n$ for a fixed constant $c>0$, the same logarithmic analysis would apply, but the final function would be rescaled in the exponent. If $x$ were allowed to be negative, the monotonicity discussion would need extra care because the outer power would no longer preserve order in the same way.

Tags: logarithmic limit, asymptotic monotonicity, product sequence