Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $\beta ={\lim }_{x\to 0}\frac{e^{x^3}-(1-x^3{)}^{1/3}+\left((1-x^2{)}^{1/2}-1\right)\sin x}{x{\sin }^{2}x}$, then the value of $6\beta$ is ______.

Why Brute Force Fails

Imagine you are standing before a massive, intimidating mathematical cliff. The problem asks us to evaluate:

Your first instinct, honed by years of standard calculus, might be to reach for L'Hopital's Rule. You see the indeterminate form $\frac{0}{0}$, and you think, 'I will just differentiate the numerator and denominator.'

Stop. Take a breath. If you differentiate that numerator, you will trigger a chain-rule explosion that will consume your time and sanity.

In JEE Advanced, the test is not just about knowing the rules; it is about choosing the most elegant path. Today, we choose elegance.

The Clean-Up

Before we touch the numerator, let us simplify the battlefield. The denominator is $x{\sin }^{2}x$.

We know the fundamental limit ${\lim }_{x\to 0}\frac{\sin x}{x}=1$. By multiplying and dividing the denominator by $x^2$, we transform it into:

As $x\to 0$, the term ${\left(\frac{\sin x}{x}\right)}^2$ becomes $1^2=1$. Suddenly, the denominator is just $x^3$.

This is a massive victory. We have reduced a trigonometric nightmare into a simple polynomial power. Now, we only need to worry about the numerator up to the power of $x^3$.

The Maclaurin Strategy

Why Maclaurin series? Because we are dealing with functions like $e^{x^3}$ and binomial powers near zero. The Maclaurin series allows us to approximate these complex functions as simple polynomials.

Since our denominator is $x^3$, any term in the numerator with a power higher than $x^3$ (like $x^4,x^5,\dots$) will effectively become zero when we divide by $x^3$ and take the limit. We are looking for the 'first non-zero coefficient' of the numerator.

Let us break down the numerator piece by piece:

1. The Exponential: $e^{x^3}$. Using the expansion $e^t=1+t+\frac{t^2}{2!}+\dots$, and substituting $t=x^3$, we get $e^{x^3}\approx 1+x^3$. We stop here because the next term is $\frac{(x^3{)}^2}{2}=\frac{x^6}{2}$, which is far beyond our needs.

2. The Binomial: $(1-x^3{)}^{1/3}$. Using the binomial expansion $(1+t{)}^n\approx 1+nt$, where $n=\frac{1}{3}$ and $t=-x^3$, we get $1-\frac{1}{3}{x}^3$.

3. The Trigonometric Product: $\left((1-x^2{)}^{1/2}-1\right)\sin x$. First, $(1-x^2{)}^{1/2}\approx 1-\frac{1}{2}{x}^2$. Subtracting the $1$ leaves us with $-\frac{1}{2}{x}^2$. Now, multiply this by the expansion of $\sin x$, which is simply $x$. The result is $-\frac{1}{2}{x}^3$.

The Final Victory

Now, we assemble our pieces back into the numerator:

Look at the constants! The $1$ and the $-1$ cancel out perfectly. We are left with $x^3+\frac{1}{3}{x}^3-\frac{1}{2}{x}^3$.

Factoring out the $x^3$, we have:

Calculating the fraction:

So, our limit $\beta$ becomes:

The question asks for $6\beta$. Multiplying $\frac{5}{6}$ by $6$ gives us $5$.

We didn't just solve a problem; we navigated a complex landscape with precision. Remember this feeling—the feeling of replacing brute force with insight. That is the heart of JEE Advanced mathematics.