Animated Solution for Mathematics - Definite Integration: Let $f:(0,1)\to \mathbb{R}$ be the functions defined as $f(x)=\sqrt{n}$ if $x\in [\frac{1}{n+1},\frac{1}{n})$ where $n\in \mathbb{N}$. Let $g:(0,1)\to \mathbb{R}$ be a function such that ${\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt<g(x)<2\sqrt{x}$ for all $x\in (0,1)$. Then ${\lim }_{x\to 0}f(x)g(x)$

A Journey into Limits

Welcome, future IITians! Today, we are going to dissect a problem that perfectly captures the elegance of JEE Advanced mathematics. It is not just about crunching numbers; it is about visualizing the behavior of functions as they dance toward a limit.

We are tasked with finding ${\lim }_{x\to 0}f(x)g(x)$, where $f(x)$ is a step function and $g(x)$ is trapped in an inequality. When you see a function 'trapped' like this, your mathematical instinct should immediately scream: The Squeeze Theorem!

Decoding the Mystery of $f(x)$

Let us first demystify $f(x)$. We are given $f(x)=\sqrt{n}$ for $x\in [\frac{1}{n+1},\frac{1}{n})$.

If $x$ is in this range, then $\frac{1}{n+1}\le x<\frac{1}{n}$. Taking the reciprocal, we get $n<\frac{1}{x}\le n+1$. This tells us that $n$ is the greatest integer less than or equal to $\frac{1}{x}$, or mathematically, $n=\lfloor \frac{1}{x}\rfloor$.

Now, we use the fundamental property of the floor function:

Since $f(x)=\sqrt{n}$, we simply take the square root of this inequality to bound our function: We have successfully caged $f(x)$!

The Squeeze Setup

We are given that ${\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt<g(x)<2\sqrt{x}$. To find the limit of $f(x)g(x)$, we multiply this inequality by our bounds for $f(x)$.

Since $f(x)$ is always positive, the inequality signs stay put. We now have a compound inequality where $f(x)g(x)$ is sandwiched between a lower bound $L(x)$ and an upper bound $U(x)$.

The Squeeze Theorem tells us that if ${\lim }_{x\to 0}L(x)={\lim }_{x\to 0}U(x)=L$, then the limit of our target function must also be $L$.

The Integral Challenge

Now for the heavy lifting. Let us evaluate the lower bound $L(x)=f(x)\cdot {\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt$. The integral part is the real challenge.

Let $I=\int \sqrt{\frac{1-t}{t}}dt$. We use the substitution $t={\sin }^2\theta$, which implies $dt=2\sin \theta \cos \theta d\theta$.

The integrand becomes $\sqrt{\frac{1-{\sin }^2\theta }{{\sin }^2\theta }}=\frac{\cos \theta }{\sin \theta }$. Multiplying by $dt$, we get:

Using the identity $2{\cos }^2\theta =1+\cos (2\theta )$, the integral becomes $\theta +\frac{\sin (2\theta )}{2}+C$. Converting back to $t$, we get $\arcsin (\sqrt{t})+\sqrt{t}\sqrt{1-t}$.

Evaluating this from $x^2$ to $x$ gives us our lower bound expression. As $x\to 0$, this expression, when multiplied by $f(x)$, converges to 2.

The Grand Finale

We have evaluated the upper bound limit, which is 2, and the lower bound limit, which is also 2. By the Squeeze Theorem, the function $f(x)g(x)$ is forced to converge to 2 as $x\to 0$.

This problem is a beautiful reminder that even the most complex-looking expressions can be tamed with the right tools—the floor function property, trigonometric substitution, and the powerful Squeeze Theorem. Keep practicing, and you will soon see these patterns everywhere!