Animated Solution for Mathematics - Differentiation: $f^{\prime }(0)={\lim }_{n\to \infty }nf(1/n)$ and $f(0)=0$. Using this find ${\lim }_{n\to \infty }\left((n+1)\frac{2}{\pi }{\cos }^{-1}\left(\frac{1}{n}\right)-n\right)\cdot \left|{\cos }^{-1}\frac{1}{n}<\frac{\pi }{2}\right|.$

Welcome, fellow traveler on the path to JEE mastery. When you first look at this problem, it is natural to feel a surge of intimidation.

You see a limit involving $n$ approaching infinity, an inverse cosine function, and a structure that seems to defy standard algebraic simplification. But here is the secret of the elite: complexity is often just a mask worn by a simple, elegant truth. Our goal today is to unmask this problem.

The Detective Work

We are given a definition: $f^{\prime }(0)={\lim }_{n\to \infty }nf(\frac{1}{n})$. At first glance, this looks like a strange, arbitrary formula.

But let us pause and think like mathematicians. Recall the definition of a derivative from first principles:

If we perform a simple substitution, letting $h=\frac{1}{n}$, then as $n\to \infty$, $h$ must approach $0$. If we assume $f(0)=0$, the expression becomes ${\lim }_{n\to \infty }\frac{f(1/n)-0}{1/n}$, which is exactly $nf(\frac{1}{n})$.

Do you see it? The problem is not asking us to solve a limit; it is asking us to recognize a derivative in disguise. We are not doing brute-force calculus; we are doing pattern recognition.

The Transformation

Now, look at our target limit: $L={\lim }_{n\to \infty }[(n+1)\frac{2}{\pi }{\cos }^{-1}(\frac{1}{n})-n]$. To make this match our derivative definition, we need an $n$ outside the bracket. Let us factor it out:

By simplifying the fraction $\frac{n+1}{n}$ to $(1+\frac{1}{n})$, we arrive at a beautiful structure. If we define our function as $f(x)=(1+x)\frac{2}{\pi }{\cos }^{-1}(x)-1$, where $x=\frac{1}{n}$, then our limit is simply $f^{\prime }(0)$.

We have successfully transformed a terrifying limit into a standard differentiation problem.

The Calculus

Now, we must differentiate $f(x)=\frac{2}{\pi }(1+x){\cos }^{-1}(x)-1$. We apply the product rule: $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$. Let $u=(1+x)$ and $v={\cos }^{-1}(x)$.

The derivative of $u$ is $1$. The derivative of $v$ is $\frac{-1}{\sqrt{1-x^2}}$. Putting it together:

I know this looks messy, but do not panic. We only need the value at $x=0$. When we substitute $x=0$, the expression collapses with satisfying simplicity. ${\cos }^{-1}(0)$ is $\frac{\pi }{2}$, and the term $\frac{-1}{\sqrt{1-0^2}}$ becomes $-1$.

The Victory

Finally, we evaluate the expression:

Distributing the $\frac{2}{\pi }$, we get $1-\frac{2}{\pi }$, which simplifies to $\frac{\pi -2}{\pi }$.

Take a moment to appreciate this. We started with a complex limit that seemed to require advanced techniques, but by identifying the underlying structure of the derivative, we reduced it to a simple product rule application.

This is the essence of JEE Advanced physics and mathematics: it is not about how much you can calculate, but how clearly you can see the underlying structure. Keep practicing this vision, and you will become unstoppable.