Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to 0}\frac{x\tan 2x-2x\tan x}{(1-\cos 2x{)}^2}$ is

The expression we are evaluating is:

Plugging in $x=0$ yields the indeterminate form $\frac{0}{0}$. While L'Hopital's Rule is a standard approach, the squared denominator suggests that repeated differentiation will lead to an algebraic nightmare. Instead, we will use the Taylor Series expansion to dissect the problem with surgical precision.

The denominator is $(1-\cos 2x{)}^2$. Using the trigonometric identity $1-\cos 2x=2{\sin }^{2}x$, we can rewrite the expression as:

As $x\to 0$, we know that $\sin x\approx x$. Therefore, the denominator behaves asymptotically as:

This serves as our "North Star." For the limit to be a finite, non-zero value, the numerator must also behave like $x^4$ as $x\to 0$.

Now, we focus on the numerator: $N=x\tan 2x-2x\tan x$. We utilize the Taylor expansion $\tan \theta \approx \theta +\frac{{\theta }^3}{3}$.

Expanding the first term, $x\tan 2x$, with $\theta =2x$:

Expanding the second term, $2x\tan x$, with $\theta =x$:

Subtracting the second term from the first, we observe the following:

The $2x^2$ terms vanish completely, revealing the underlying structure of the function. We are left with:

We now combine our simplified numerator and denominator to evaluate the limit:

The $x^4$ terms cancel out, leaving us with the final result:

By avoiding the brute force of L'Hopital's Rule, we have navigated the problem with efficiency. In JEE Advanced, the smartest path is often the one that simplifies the expression before you begin calculating.