Animated Solution for Mathematics - Limits, Continuity and Differentiability: If the function $f(x)=\frac{\sin 3x+\alpha \sin x-\beta \cos 3x}{x^3},x\in \mathbb{R}$, is continuous at $x=0$, then $f(0)$ is equal to :

Imagine you are standing at the edge of a mathematical cliff at $x=0$. The function $f(x)=\frac{\sin 3x+\alpha \sin x-\beta \cos 3x}{x^3}$ looks intimidating. It is a ratio of trigonometric functions, and as $x$ approaches zero, the denominator $x^3$ shrinks to nothingness.

In the world of limits, a denominator approaching zero is a red flag—it usually signals a vertical asymptote, a place where the function explodes to infinity. But here, we are told the function is continuous. This is our golden ticket.

It tells us that the numerator must be 'hiding' enough zeros to cancel out that $x^3$ in the denominator. This leaves us with a finite, elegant value at the origin.

Instead of wrestling with trigonometric identities, let us use the most powerful tool in our arsenal: the Maclaurin series expansion. Think of these series as a way to 'zoom in' on the function near $x=0$.

We know that the standard expansions are:

By replacing the complex trig functions with these simple polynomials, we strip away the mystery. Substituting these into our numerator, we get:

Now, let us organize this expression. We group the terms by the power of $x$:

Look closely at this expression. We have a constant term, an $x$ term, an $x^2$ term, and an $x^3$ term. If any of the terms with powers lower than $x^3$ were non-zero, the limit would be undefined.

For example, if the constant term $-\beta$ were not zero, we would have a term like $-\beta /x^3$, which shoots to infinity as $x\to 0$. To ensure continuity, we must force these 'troublemakers' to vanish by setting their coefficients to zero:

1. The constant term: $-\beta =0\Rightarrow \beta =0$. 2. The coefficient of $x$: $3+\alpha =0\Rightarrow \alpha =-3$. 3. The coefficient of $x^2$: $\frac{9\beta }{2}=0$. Since $\beta =0$, this is automatically satisfied.

With $\alpha =-3$ and $\beta =0$, the lower-order terms vanish, leaving us with only the $x^3$ term in the numerator. The function simplifies beautifully to:

As $x^3$ cancels out, we are left with the limit itself. Substituting our value $\alpha =-3$:

And there it is. The chaos of the trigonometric functions has collapsed into a single, clean integer. You have successfully navigated the limit, tamed the denominator, and found the value that keeps the function continuous.

The final result is $-4$. Remember, in JEE Advanced, the most complex-looking problems often have the most symmetrical, elegant solutions. Keep looking for the pattern, and the math will always reveal its secrets.