Animated Solution for Mathematics - Limits, Continuity and Differentiability: If ${\lim }_{x\to 0}\frac{\alpha x{e}^x-\beta {\log }_e(1+x)+\gamma x^2{e}^{-x}}{x{\sin }^{2}x}=10$, $\alpha ,\beta ,\gamma \in \mathbb{R}$, then the value of $\alpha +\beta +\gamma$ is

A Journey Through Taylor Series

Welcome, future engineers. Today, we are not just solving a limit problem; we are performing a surgical operation on a mathematical expression.

When you see a limit like

it is easy to feel overwhelmed. You see transcendental functions like $e^x$ and ${\log }_e(1+x)$ mixed with polynomials, but in the world of JEE Advanced, we don't panic; we analyze.

The Denominator's Secret

Before we touch the numerator, we must understand the 'stage' upon which this drama unfolds: the denominator, $x{\sin }^{2}x$.

As $x$ approaches $0$, we know that $\sin x\approx x$. Therefore, the denominator behaves as follows:

This is the most critical realization of the entire problem. The denominator is essentially an $x^3$ term, which implies the numerator must also be an $x^3$ polynomial for the limit to be a finite, non-zero value.

The Taylor Series Toolkit

Now, we bring in the heavy artillery: Taylor Series expansions. These allow us to convert complex functions into simple polynomials:

By substituting these into our numerator, we 'unmask' the functions to see their true polynomial nature:

1. $\alpha x{e}^x=\alpha x(1+x+\frac{x^2}{2}+\dots )=\alpha x+\alpha x^2+\frac{\alpha x^3}{2}$

2. $-\beta {\log }_e(1+x)=-\beta (x-\frac{x^2}{2}+\frac{x^3}{3}-\dots )=-\beta x+\frac{\beta x^2}{2}-\frac{\beta x^3}{3}$

3. $\gamma x^2{e}^{-x}=\gamma x^2(1-x+\dots )=\gamma x^2-\gamma x^3$

The Algebraic Siege

Now, we group these terms by their powers of $x$. We collect the coefficients for $x$, $x^2$, and $x^3$:

Coefficient of $x$: $(\alpha -\beta )$ Coefficient of $x^2$: $(\alpha +\frac{\beta }{2}+\gamma )$ * Coefficient of $x^3$: $(\frac{\alpha }{2}-\frac{\beta }{3}-\gamma )$

For the limit to be $10$, the terms with $x$ and $x^2$ must vanish. This gives us a system of equations:

The Final Victory

With the lower-order terms eliminated, the limit simplifies to the ratio of the $x^3$ coefficients:

Substituting $\beta =\alpha$ and $\gamma =-\frac{3\alpha }{2}$ into the equation:

Consequently, $\beta =6$ and $\gamma =-9$. The final sum is:

You have successfully navigated the complexity and arrived at the truth. This is the essence of JEE Advanced mathematics—the elegant mastery of structure.