Animated Solution for Mathematics - Limits, Continuity and Differentiability: If ${\lim }_{x\to 0}\frac{a{e}^x-b\cos x+c{e}^{-x}}{x\sin x}=2$, then $a+b+c$ is equal to ______

A Journey into Taylor Series

Imagine you are standing on the edge of a mathematical cliff, looking down at the expression ${\lim }_{x\to 0}\frac{a{e}^x-b\cos x+c{e}^{-x}}{x\sin x}=2$. It looks intimidating, but in the world of JEE Advanced, complexity is often just a mask for elegance. Let's peel back that mask together.

The Anatomy of the Denominator

Before we touch the numerator, we must understand the ground we stand on. The denominator is $x\sin x$. As $x$ dances toward zero, $\sin x$ behaves remarkably like $x$ itself.

This is the fundamental small-angle approximation: $\sin x\approx x$. Therefore, our denominator $x\sin x$ behaves like $x\cdot x=x^2$.

This is our anchor. It tells us that the entire expression is governed by the $x^2$ behavior. If the numerator doesn't match this $x^2$ rhythm, the whole thing will either collapse to zero or explode to infinity.

The Taylor Series Toolkit

We could try L'Hopital's rule, but that path is fraught with tedious differentiation. Instead, let's use the scalpel of calculus: the Taylor series. We expand our functions near $x=0$ to see their true nature:

These aren't just random polynomials; they are the DNA of these functions. By substituting these into our numerator $a{e}^x-b\cos x+c{e}^{-x}$, we can see exactly what is happening under the hood.

The Algebraic Cleanup

Let's perform the substitution carefully. We have:

Now, we group the terms by their powers of $x$. This is where the magic happens:

- Constant terms ($x^0$): $(a-b+c)$ - Linear terms ($x^1$): $(a-c)x$ - Quadratic terms ($x^2$): $\left(\frac{a}{2}+\frac{b}{2}+\frac{c}{2}\right)x^2$

Our numerator is now $N\approx (a-b+c)+(a-c)x+\left(\frac{a+b+c}{2}\right)x^2$.

The Logic of Finiteness

We know the limit is $2$. If the constant term $(a-b+c)$ were not zero, the limit would be $\frac{\text{constant}}{x^2}$, which is undefined.

If the linear term $(a-c)$ were not zero, the limit would be $\frac{\text{linear}}{x^2}=\frac{1}{x}$, which is also undefined. Therefore, for the limit to exist, these terms must vanish.

We set $a-b+c=0$ and $a-c=0$. This leaves us with only the $x^2$ term!

The Final Cancellation

With the lower terms gone, our limit simplifies beautifully:

The $x^2$ terms cancel out, leaving us with the simple equation $\frac{a+b+c}{2}=2$. A quick multiplication by $2$ gives us the final result:

Isn't that satisfying? We didn't need to find $a$, $b$, and $c$ individually. We just needed to understand the structure of the limit. Keep this in mind: when you see limits, don't just calculate—visualize the behavior.