Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)={\lim }_{n\to \infty }{\sum }_{r=0}^n(\frac{\tan (x/2^{r+1})+{\tan }^3(x/2^{r+1})}{1-{\tan }^2(x/2^{r+1})}).$ Then ${\lim }_{x\to 0}\frac{e^x-e^{f(x)}}{x-f(x)}$ is equal to

My dear student, I know exactly what you felt when you first laid eyes on this problem. It looks like a nightmare, doesn't it? A limit of a summation, trigonometric functions raised to powers, and an exponential limit all wrapped into one.

But here is the secret of the JEE Advanced: the most intimidating problems are often the most elegant. They are designed to test your ability to see through the noise. Let us peel back the layers of this monster together.

Deconstructing the General Term

We start by focusing on the general term of the summation:

Let $\theta =x/2^{r+1}$. The expression becomes:

Now, remember your fundamental identities? Since $1+{\tan }^2(\theta )={\sec }^2(\theta )$, our term becomes:

Converting to sine and cosine, we have $\tan (\theta )=\sin (\theta )/\cos (\theta )$ and ${\sec }^2(\theta )=1/{\cos }^2(\theta )$. The denominator $1-{\tan }^2(\theta )$ becomes $({\cos }^2(\theta )-{\sin }^2(\theta ))/{\cos }^2(\theta )$.

When you simplify this complex fraction, the ${\cos }^2(\theta )$ terms cancel out beautifully, leaving us with:

Since ${\cos }^2(\theta )-{\sin }^2(\theta )=\cos (2\theta )$, we arrive at:

The Telescoping Magic

Now, here is where the real magic happens. We want to express $T_r$ as a difference. Consider the identity:

Taking the common denominator, we get:

This is exactly our $T_r$! Substituting back $\theta =x/2^{r+1}$, we get:

This is a perfect telescoping series. When we sum this from $r=0$ to $n$, the terms cancel out:

Everything in the middle vanishes, leaving only:

The Final Limit

As $n\to \infty$, the term $\tan (x/2^{n+1})$ approaches $\tan (0)=0$. Thus, $f(x)=\tan (x)$.

Now, we evaluate the original limit:

This is a $0/0$ form. Instead of using L'Hopital's rule, let us factor out $e^{\tan (x)}$:

We know that ${\lim }_{u\to 0}\frac{e^u-1}{u}=1$. Here, $u=x-\tan (x)$, which approaches $0$ as $x\to 0$.

Therefore, the limit is $e^{\tan (0)}\cdot 1=e^0\cdot 1=1$.

The monster is defeated. The final answer is 1. Always remember, in JEE, look for the structure, trust your identities, and never lose hope.