If lim(x→0) (αeˣ + βe⁻ˣ + γsin x) / (x sin² x) = 2/3, where α, β, γ ∈ R, then which of the following

Q: Why do we just ignore terms higher than $x^3$ in the Taylor expansion?

That is a fantastic question! We are taking the limit as $x \to 0$. Any term like $x^4$ or $x^5$ in the numerator, when divided by the denominator $x^3$, will result in terms like $x$ or $x^2$. As $x$ approaches zero, these terms simply vanish to zero. They contribute nothing to the final limit, so we discard them to keep our algebra clean and focused.

Q: Can I use L'Hopital's Rule instead of Taylor Series?

Technically, yes. But in the JEE Advanced arena, time and precision are your currency. Differentiating a complex expression three times with three unknown variables ($\alpha, \beta, \gamma$) is a recipe for a calculation error. Taylor Series acts like a surgical scalpel—it cuts through the complexity and gives you the coefficients directly. It is the 'pro' way to solve this.

Q: Why must the coefficients of $x^0, x^1,$ and $x^2$ be zero?

Think of the limit as a balance. If the numerator had a constant term (like $5$), the limit would be $\frac{5}{x^3}$, which blows up to infinity as $x \to 0$. If it had an $x$ term, it would be $\frac{x}{x^3} = \frac{1}{x^2}$, which also blows up. To get a finite, non-zero value like $\frac{2}{3}$, the 'noise' of the lower powers must be silenced. Setting them to zero is the only way to ensure the limit converges.

Question

If $\lim_{x 	o 0} \frac{\alpha e^x + \beta e^{-x} + \gamma \sin x}{x \sin^2 x} = \frac{2}{3}$, where $\alpha, \beta, \gamma \in \mathbb{R}$, then which of the following is NOT correct ?

Options:

(A) $\alpha^2 + \beta^2 + \gamma^2 = 6$

(B) $\alpha\beta + \beta\gamma + \gamma\alpha + 1 = 0$

(C) $\alpha\beta^2 + \beta\gamma^2 + \gamma\alpha^2 + 3 = 0$

(D) $\alpha^2 - \beta^2 + \gamma^2 = 4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

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

$$\lim_{x 	o 0} \frac{\alpha e^x + \beta e^{-x} + \gamma \sin x}{x \sin^2 x} = \frac{2}{3}$$

most students see a wall of variables. I want you to see a puzzle waiting to be solved. This problem is a classic JEE Advanced favorite because it tests your ability to look past the surface and understand the 'order' of functions.

### The Denominator's Secret

Before we touch the numerator, we must understand the ground we are standing on. The denominator is $x \sin^2 x$. In the world of limits, as $x 	o 0$, the function $\sin x$ is not just a trigonometric function; it is effectively $x$.

We replace $\sin x$ with $x$, turning our denominator into $x \cdot (x)^2$, which is $x^3$.

Why does this matter? Because the denominator tells us the 'scale' of the problem. It tells us that the numerator must also be of the order $x^3$ for the limit to be finite. If the numerator had a constant term, the limit would be infinite; if it had an $x$ term, it would still be infinite. The denominator is the gatekeeper, and it demands that we eliminate all lower-order chaos.

### The Taylor Series Weapon

Now, we bring out the heavy artillery: the Taylor Series. We need to express our functions as polynomials. We know that:

$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots$$

$$e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \dots$$

$$\sin x = x - \frac{x^3}{6} + \dots$$

I know what you are thinking: 'Do I really need to memorize these?' Yes, but more importantly, you need to understand them. These series are just approximations. As $x$ gets smaller and smaller, these polynomials become indistinguishable from the original functions. We substitute these into our numerator, and suddenly, the expression transforms from a scary fraction into a structured polynomial.

### The Constraint Game

This is where the magic happens. We substitute the series into the numerator:

$$	ext{Numerator} = \alpha\left(1 + x + \frac{x^2}{2} + \frac{x^3}{6}ight) + \beta\left(1 - x + \frac{x^2}{2} - \frac{x^3}{6}ight) + \gamma\left(x - \frac{x^3}{6}ight)$$

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

$$(\alpha + \beta) + (\alpha - \beta + \gamma)x + \left(\frac{\alpha + \beta}{2}ight)x^2 + \left(\frac{\alpha - \beta - \gamma}{6}ight)x^3 + \dots$$

Remember our gatekeeper? The denominator is $x^3$. If any of the coefficients for $x^0, x^1,$ or $x^2$ are non-zero, the limit will explode to infinity. To keep the limit at $\frac{2}{3}$, we must force these coefficients to be zero.

1. For $x^0$: $\alpha + \beta = 0 \implies \beta = -\alpha$
2. For $x^1$: $\alpha - \beta + \gamma = 0$. Substituting $\beta = -\alpha$, we get $\alpha - (-\alpha) + \gamma = 0$, which means $2\alpha + \gamma = 0$, or $\gamma = -2\alpha$.

### The Final Reveal

With the lower powers silenced, only the $x^3$ term remains to face the denominator. The limit becomes:

$$\lim_{x 	o 0} \frac{\left(\frac{\alpha - \beta - \gamma}{6}ight)x^3}{x^3} = \frac{\alpha - \beta - \gamma}{6} = \frac{2}{3}$$

Now, we substitute our relationships ($\beta = -\alpha$ and $\gamma = -2\alpha$) into this equation:

$$\frac{\alpha - (-\alpha) - (-2\alpha)}{6} = \frac{2}{3}$$

$$\frac{4\alpha}{6} = \frac{2}{3} \implies \frac{2\alpha}{3} = \frac{2}{3} \implies \mathbf{\alpha = 1}$$

With $\alpha = 1$, the rest falls like dominoes: $\mathbf{\beta = -1}$ and $\mathbf{\gamma = -2}$.

### The Takeaway

We didn't just 'solve' a problem; we navigated a landscape of indeterminate forms using the most elegant tool in calculus. Whenever you see a limit involving exponentials or trigonometric functions, don't panic. Expand, group, constrain, and solve. That is the rhythm of the JEE Advanced. Keep practicing, keep questioning, and keep falling in love with the logic.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{x \to 0} \frac{α e ^{x} + β e ^{- x} + γ s i n x}{x s i n ^{2} x} = \frac{2}{3}$ , where $α, β, γ \in R$ , then which of the following is NOT correct ?

Select Answer:

Visualized Solution

Analyzing the Limit Form

Taylor Series Expansion

Substituting Expansions

Grouping Powers of $x$

Condition for Finite Limit

Eliminating the Constant Term

Eliminating the Linear Term

Consistency Check

Evaluating the Limit Value

Solving for $α$

Finding $β$ and $γ$

Checking Options A and B

Checking Options C and D

Summary and Takeaway

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Denominator's Secret

The Taylor Series Weapon

The Constraint Game

The Final Reveal

The Takeaway

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