Animated Solution for Mathematics - Limits, Continuity and Differentiability: If ${\lim }_{x\to 0}\left\{(1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}+\cos \frac{x^2}{2}\cos \frac{x^2}{4})\right\}=2^k$, then the value of $k$ is

Welcome, fellow traveler in the world of mathematics. Today, we are standing before a limit that, at first glance, looks like a tangled mess of trigonometric functions.

We are looking at the expression:

In the JEE Advanced arena, intimidation is just a test of your composure. Let's break this beast down, piece by piece, and find the soul of this problem.

When you see a four-term expression like the numerator, your first instinct should be to look for structure. It is not just a random collection of terms; it is a hidden product.

Let's group the first two terms and the last two terms:

Do you see it? The binomial $(1-\cos \frac{x^2}{2})$ is common to both parts. By factoring it out, we transform a complex addition-subtraction problem into a clean multiplication:

We have just simplified the numerator from a chaotic four-term expression into an elegant product of two binomials. This is the first step in turning a 'problem' into a 'solution'.

Now, we have terms in the form $1-\cos \theta$. In the world of limits as $x\to 0$, cosines are difficult because they approach $1$, leading to indeterminate forms.

Sines, however, are our best friends because of the standard limit ${\lim }_{u\to 0}\frac{\sin u}{u}=1$. We use the half-angle identity $1-\cos \theta =2{\sin }^2\frac{\theta }{2}$ to bridge this gap.

For our first factor, $\theta =\frac{x^2}{2}$, so:

For our second factor, $\theta =\frac{x^2}{4}$, so:

Our numerator is now $4{\sin }^2\frac{x^2}{4}{\sin }^2\frac{x^2}{8}$.

We are now at the final hurdle. We have:

To use the standard limit, we need the denominator to match the argument of the sine function. For ${\sin }^2\frac{x^2}{4}$, we need $(\frac{x^2}{4}{)}^2$ in the denominator. For ${\sin }^2\frac{x^2}{8}$, we need $(\frac{x^2}{8}{)}^2$.

We multiply and divide by these terms to balance the expression:

The $x^4\cdot x^4$ in the numerator perfectly cancels the $x^8$ in the denominator. We are left with:

We found that the limit is $\frac{1}{256}$. Since $256=2^8$, we can write this as $2^{-8}$.

The problem states the limit is $2^k$, so $2^k=2^{-8}$, which means $k=-8$.

We have conquered the beast! Remember, the complexity of a problem is often just a mask for a simple, elegant truth waiting to be uncovered. Keep practicing, keep questioning, and most importantly, keep falling in love with the process.