Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to 0}\frac{{\sin }^2(\pi {\cos }^{4}x)}{x^4}$ is equal to :

A Journey into Calculus

Limits are the heartbeat of calculus. They allow us to peer into the behavior of functions as they approach a point, even when the function itself might be undefined or indeterminate at that exact location.

Today, we are going to tackle a problem that looks intimidating at first glance:

It is a classic JEE Advanced problem that tests not just your algebraic skills, but your ability to see the underlying structure of trigonometric functions.

The Indeterminate Trap

Whenever you face a limit, your first instinct should always be to check the form. Let's substitute $x=0$ into our expression.

The denominator, $x^4$, clearly approaches $0$. In the numerator, we have ${\sin }^2(\pi {\cos }^{4}0)$. Since $\cos 0=1$, this becomes ${\sin }^2(\pi \cdot 1^4)={\sin }^2(\pi )=0$.

We have arrived at a $\frac{0}{0}$ indeterminate form. This is our green light to start simplifying.

The Trigonometric Bridge

We have a ${\sin }^2\theta$ term, which is often difficult to handle directly in a limit. We need to linearize it.

The identity ${\sin }^2\theta =\frac{1-\cos (2\theta )}{2}$ is our best friend here. By setting $\theta =\pi {\cos }^{4}x$, our expression transforms into:

Now, we are dealing with a cosine function, which is much easier to manipulate using standard limits.

The Algebraic Dance

We still have that ${\cos }^{4}x$ inside the cosine. Since we are working with $x\to 0$, we want to move toward sine functions, because we know ${\lim }_{x\to 0}\frac{\sin x}{x}=1$.

We use the identity ${\cos }^{2}x=1-{\sin }^{2}x$, which means ${\cos }^{4}x=(1-{\sin }^{2}x{)}^2$. Expanding this, we get $1-2{\sin }^{2}x+{\sin }^{4}x$.

Multiplying by $2\pi$, the argument of our cosine becomes $2\pi -4\pi {\sin }^{2}x+2\pi {\sin }^{4}x$.

Here is where the magic happens. Using the periodicity property $\cos (2\pi -\theta )=\cos \theta$, we can ignore the $2\pi$ and simplify the argument to $4\pi {\sin }^{2}x-2\pi {\sin }^{4}x$.

Our limit now looks like:

The Final Convergence

We are almost there! We use the identity $1-\cos \alpha =2{\sin }^2(\frac{\alpha }{2})$.

Applying this to our numerator, we get $2{\sin }^2(2\pi {\sin }^{2}x-\pi {\sin }^{4}x)$. The $2$ in the numerator cancels with the $2$ in the denominator, leaving us with:

To finish, we use the standard limit ${\lim }_{u\to 0}\frac{\sin u}{u}=1$. We multiply and divide by the square of the argument, $(2\pi {\sin }^{2}x-\pi {\sin }^{4}x{)}^2$.

The limit becomes:

Factoring out $\pi {\sin }^{2}x$, we get:

Since $\frac{\sin x}{x}\to 1$, this simplifies to ${\pi }^2\cdot 1^4\cdot (2-0{)}^2=4{\pi }^2$.

And there it is! Through careful application of identities and standard limits, we have unraveled the complexity to find the answer: $4{\pi }^2$.