Animated Solution for Mathematics - Definite Integration: Let $f(x)={\lim }_{n\to \infty }{\left(\frac{n^n(x+n)(x+\frac{n}{2})...(x+\frac{n}{n})}{n!(x^2+n^2)(x^2+\frac{n^2}{4})...(x^2+\frac{n^2}{n^2})}\right)}^{x/n}$, for all $x>0$. Then

Taming the Limit

My dear student, look at the expression on your screen. It is designed to intimidate. It has a limit as $n\to \infty$, a massive product of terms, and an exponent that depends on $x$.

In the world of JEE Advanced, we do not run from monsters; we dissect them. This problem is a beautiful example of how complex structures often hide elegant, simple truths.

The Logarithmic Transformation

Whenever you encounter a limit of a product raised to a variable power, your first instinct should be to take the natural logarithm. The logarithm is the great simplifier; it turns products into sums and brings exponents down to the ground level.

Let $f(x)$ be our expression. By taking the natural logarithm, we get:

Suddenly, the exponent $\frac{x}{n}$ is no longer a threat; it is just a coefficient. The product is now inside the logarithm, waiting to be broken down into a sum of logarithms.

The Riemann Sum Revelation

Now, let us organize the terms. We know that $n!={\prod }_{k=1}^{n}k$ and $n^n={\prod }_{k=1}^{n}n$. By grouping these, we can rewrite the entire product as:

When we simplify the general term of this product, we get $\frac{nx+\frac{n^2}{k}}{k{x}^2+\frac{n^2}{k}}$. If we divide both the numerator and the denominator by $\frac{n^2}{k}$, we obtain:

This is the moment of clarity! We have successfully expressed the term as a function of $\frac{k}{n}$. The expression becomes:

This is the classic signature of a Riemann sum. We are looking at ${\lim }_{n\to \infty }\frac{1}{n}{\sum }_{k=1}^{n}g(\frac{k}{n})$, which is the definition of the definite integral ${\int }_0^{1}g(y)dy$.

The Calculus of Elegance

With the Riemann sum identified, our limit transforms into a beautiful integral:

To make this even cleaner, we use a substitution. Let $t=xy$. Then $dt=xdy$, and the limits change from $0$ to $x$. The $x$ outside the integral cancels out perfectly with the $x$ from the differential $dy$.

We are left with:

This is the heart of the problem. We have reduced a terrifying limit to a simple integral.

The Final Analysis

Now, the question asks us to analyze the behavior of $f(x)$. To do this, we need the derivative $f^{\prime }(x)$. We use the Leibniz Rule (or the Fundamental Theorem of Calculus) to differentiate the integral:

This gives us:

To find the critical points, we set $f^{\prime }(x)=0$, which implies $\ln \left(\frac{1+x}{1+x^2}\right)=0$, or $\frac{1+x}{1+x^2}=1$. Solving this quadratic equation $1+x=1+x^2$ gives us $x=0$ or $x=1$.

Since $x>0$, our critical point is $x=1$. By testing values, we see that for $x<1$, the derivative is positive (the function increases), and for $x>1$, the derivative is negative (the function decreases). Thus, $f(1)$ is the absolute maximum. We have conquered the monster!