Animated Solution for Mathematics - Limits, Continuity and Differentiability: For $\alpha ,\beta ,\gamma ,\in \mathbb{R}$, if ${\lim }_{x\to 0}\frac{x^2\sin \alpha x+(\gamma -1)e^{x^2}}{\sin 2x-\beta x}=3$, then $\beta +\gamma -\alpha$ is equal to:

Welcome, future engineer. Today, we are not just solving a limit problem; we are performing a delicate balancing act. When you look at a limit like

it is easy to feel overwhelmed by the variables $\alpha$, $\beta$, and $\gamma$. But I want you to see this not as a chaotic equation, but as a puzzle of symmetry. In the world of JEE Advanced, limits are rarely about brute force; they are about understanding the 'DNA' of a function near a specific point.

Imagine you are looking at a complex curve through a microscope. As $x$ approaches $0$, the function behaves like a simple polynomial. This is the magic of Taylor series. We don't need to know the entire function; we only need to know how it starts.

We recall our fundamental expansions:

These are our tools. We are going to replace the complex trigonometric and exponential terms with these simple polynomials. This is the moment where the chaos begins to organize itself.

Let's look at the numerator: $x^2\sin \alpha x+(\gamma -1)e^{x^2}$.

First, the sine term. We substitute $\alpha x$ into our sine expansion:

Notice how the power of $x$ jumped from $2$ to $3$ because of that $x^2$ multiplier. Keep that in mind.

Next, the exponential term. We substitute $x^2$ into the $e^u$ expansion:

When we combine these, our numerator looks like this: $(\gamma -1)+(\gamma -1)x^2+\alpha x^3+\dots$. This is the 'DNA' of our numerator. It has a constant term, an $x^2$ term, and an $x^3$ term.

Now, let's look at the denominator: $\sin 2x-\beta x$.

Using our expansion for $\sin 2x$, we get:

Subtracting $\beta x$, we get $(2-\beta )x-\frac{4}{3}{x}^3+\dots$.

Here is the critical insight. The denominator starts with an $x$ term. But wait! Our numerator starts with a constant term $(\gamma -1)$. If we divide a constant by an $x$ term as $x\to 0$, the result is infinity.

But the problem tells us the limit is $3$. This is a contradiction! The only way to resolve this is to force the lower-order terms to vanish.

For the limit to be finite, the numerator and denominator must 'start' at the same power of $x$.

1. The $x$ term: The denominator has an $x$ term with coefficient $(2-\beta )$. If this survives, the numerator must have an $x$ term. It doesn't. So, we must kill it. Set $2-\beta =0$, which gives $\beta =2$.

2. The constant and $x^2$ terms: Now that the $x$ term is gone, the denominator starts at $x^3$ with coefficient $-\frac{4}{3}$. For the limit to be $3$, the numerator must also start at $x^3$. This means the constant $(\gamma -1)$ and the $x^2$ coefficient $(\gamma -1)$ must be zero. Thus, $\gamma -1=0$, which gives $\gamma =1$.

With $\beta =2$ and $\gamma =1$, the limit simplifies beautifully to the ratio of the coefficients of $x^3$:

We have found our values: $\alpha =-4$, $\beta =2$, and $\gamma =1$. The final calculation, $\beta +\gamma -\alpha$, becomes $2+1-(-4)=7$.

See? It wasn't about memorizing formulas; it was about understanding the hierarchy of powers. You have successfully navigated the limit. Keep this mindset—always look for the dominant term, and the rest will fall into place.