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

A Journey Through ${\lim }_{x\to 0}\frac{\cos (\sin x)-\cos x}{x^4}$

Welcome, my dear student, to a masterclass in limits. Today, we are going to dissect a problem that often strikes fear into the hearts of aspirants:

It looks intimidating, doesn't it? A nested trigonometric function, a difference of cosines, and a fourth-power denominator. But fear not. In the world of JEE Advanced, complexity is merely an invitation to use the right tools. Let us peel back the layers of this problem together.

The Indeterminate Trap

Every limit problem begins with a simple test: direct substitution. If we plug in $x=0$, we get $\cos (\sin 0)-\cos 0=\cos (0)-1=1-1=0$ in the numerator, and $0^4=0$ in the denominator.

We have hit the classic $\frac{0}{0}$ indeterminate form. This is our signal to stop, breathe, and prepare for some algebraic surgery. We cannot simply 'solve' this; we must transform it.

The Trigonometric Transformation

Look closely at the numerator: $\cos (\sin x)-\cos x$. This is a difference of two cosines. Whenever you see $\cos A-\cos B$, your mind should immediately jump to the product-to-sum identity:

By applying this, we convert a subtraction into a multiplication. Why is this better? Because multiplication is far more 'limitable'—it allows us to isolate terms and apply standard limits.

Substituting $A=\sin x$ and $B=x$, our expression becomes:

Now, we are getting somewhere.

The Standard Limit Strategy

We know the holy grail of limits: ${\lim }_{\theta \to 0}\frac{\sin \theta }{\theta }=1$. Our goal is to force our expression into this form. We have two sine terms in the numerator.

Let us divide and multiply each by its own argument. For the first term, we multiply and divide by $\frac{x+\sin x}{2}$. For the second, we do the same with $\frac{x-\sin x}{2}$.

As $x\to 0$, these sine terms will effectively become their arguments, and the 'sine' part will vanish into the value of 1. This leaves us with:

The Taylor Series Masterclass

We are left with $\frac{x^2-{\sin }^{2}x}{x^4}$. If we try to use L'Hopital's rule here, we will be stuck in a loop of derivatives. Instead, let us use the Taylor series expansion for $\sin x=x-\frac{x^3}{6}+\dots$.

When we square this, we get:

Substituting this back into our expression, the $x^2$ terms cancel out, leaving us with $\frac{x^4}{3}$ in the numerator.

Now, look at the beauty of the final step:

The $x^4$ terms cancel out perfectly, leaving us with $\frac{1}{2}\cdot \frac{1}{3}=\frac{1}{6}$. We have conquered the beast!

The limit is $\frac{1}{6}$. Remember, in JEE, it is not about brute force; it is about recognizing the structure and applying the right tool at the right time. Keep practicing, and soon, these problems will feel like second nature.