Animated Solution for Mathematics - Limits, Continuity and Differentiability: If the function $f(x)=\frac{\cos (\sin x)-\cos x}{x^4}$ is continuous at each point in its domain and $f(0)=\frac{1}{k}$, then $k$ is ____

Imagine you are walking along a path defined by the function $f(x)=\frac{\cos (\sin x)-\cos x}{x^4}$. As you approach $x=0$, you notice a problem: the function is undefined at that exact point.

But we are told the function is continuous. In the world of calculus, this means there is a "hole" in our graph, and we need to find the exact value $f(0)=\frac{1}{k}$ that fills this hole perfectly, making the path smooth and unbroken.

Our mission is to find the value of $k$ by calculating the limit of the function as $x$ approaches zero.

When you look at the numerator, $\cos (\sin x)-\cos x$, your first instinct might be to panic or try a complex expansion. Don't! In JEE problems, whenever you see a difference of cosines, your mind should immediately jump to the sum-to-product identity:

Here, let $C=\sin x$ and $D=x$. By substituting these into our identity, the numerator transforms into:

Suddenly, the subtraction that was blocking our path has turned into a product. This is the power of trigonometric identities—they reshape the problem into something manageable.

Now, we look at our limit expression again:

We have an $x^4$ in the denominator. If we just leave it there, we are stuck. But look at the angles of our sine terms: $\frac{x+\sin x}{2}$ and $\frac{x-\sin x}{2}$.

As $x\to 0$, the first term behaves like $x$, and the second term behaves like $x^3$. This is our cue! We split $x^4$ into $x\cdot x^3$ and pair them strategically:

Let's tackle the first part, $L_1={\lim }_{x\to 0}\frac{\sin \left(\frac{x+\sin x}{2}\right)}{x}$. By multiplying and dividing by the angle $\frac{x+\sin x}{2}$, we get:

Now for the second part, $L_2={\lim }_{x\to 0}\frac{\sin \left(\frac{x-\sin x}{2}\right)}{x^3}$. Again, we balance the expression by multiplying and dividing by the angle $\frac{x-\sin x}{2}$:

This is where the Taylor series expansion of $\sin x\approx x-\frac{x^3}{6}$ saves the day. Substituting this in, we get:

We have our two pieces: $L_1=1$ and $L_2=\frac{1}{12}$. Bringing them back into our master equation:

Therefore, $k=6$. You have successfully filled the hole in the function, ensuring continuity. It is a beautiful result, isn't it? The complexity of the original expression collapses into a simple integer.