Let α, β ∈ R be such that lim (x→0) (x² sin(βx))/(αx - sin x) = 1. Then 6(α + β) equals .........

Q: Why can't I just use L'Hopital's Rule? It works for all $0/0$ forms, right?

While L'Hopital's Rule is technically valid, it is a trap in this specific problem! Differentiating $\sin(\beta x)$ multiple times will generate a complex chain of derivatives that makes the algebra prone to errors. Series expansion is the surgical tool here; it allows you to see the 'DNA' of the function near zero without the brute force of differentiation.

Q: How do I know when to stop the series expansion? Do I need to go up to $x^5$ or $x^7$?

Great question! You only need to expand until you find the first non-zero term. In this problem, the numerator starts with $x^3$ (because of $x^2 \cdot x$). Therefore, you only need to expand the denominator until you find the $x^3$ term. Any higher-order terms like $x^5$ or $x^7$ will vanish as $x \to 0$, so they don't affect the final limit.

Q: Why must the coefficient of the $x$ term in the denominator be zero?

Think about the limit as a ratio of powers. If the denominator has an $x^1$ term and the numerator only has $x^3$ terms, the expression behaves like $\frac{x^3}{x} = x^2$. As $x \to 0$, this goes to $0$. But the problem states the limit is $1$. To get a non-zero constant, the powers must match. By setting the coefficient of $x$ to zero, we 'promote' the $x^3$ term to be the new leading term, allowing the limit to be a finite constant.

Question

Let $\alpha, \beta \in R$ be such that $\lim_{x 	o 0} \frac{x^2 \sin(\beta x)}{\alpha x - \sin x} = 1$. Then $6(\alpha + \beta)$ equals .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE Advanced mastery. Today, we are not just solving a limit problem; we are uncovering the hidden geometry of functions near the origin.

When you look at the expression

$$\lim_{x 	o 0} \frac{x^2 \sin(\beta x)}{\alpha x - \sin x} = 1$$

your first instinct might be to reach for L'Hopital's Rule. But pause. In the world of competitive exams, speed and elegance are your best friends. Let us look at this through the lens of Maclaurin series, a tool that reveals the true nature of functions.

### The Gatekeeper: The Indeterminate Form

Every limit problem begins with a check. We substitute $x = 0$ into our expression.

The numerator becomes $0^2 \cdot \sin(0) = 0$. The denominator becomes $\alpha(0) - \sin(0) = 0$.

We have arrived at the $0/0$ indeterminate form. This is the gatekeeper; it tells us that the function is hiding its true value, and we must peel back the layers to see what lies beneath.

### The Microscope: Maclaurin Series

To see what is happening near $x = 0$, we use the Maclaurin series. Think of this as a microscope that zooms into the function's behavior at the origin.

We know that:

$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots = x - \frac{x^3}{6} + \dots$$

Similarly, for $\sin(\beta x)$, we replace $x$ with $\beta x$:

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

### The Balancing Act: Matching the Powers

Now, let us substitute these series into our limit. The numerator becomes:

$$x^2 \left( \beta x - \frac{\beta^3 x^3}{6} + \dots ight) = \beta x^3 - \frac{\beta^3 x^5}{6} + \dots$$

The denominator becomes:

$$\alpha x - \left( x - \frac{x^3}{6} + \dots ight) = (\alpha - 1)x + \frac{x^3}{6} - \dots$$

Here is the crucial insight: the problem tells us the limit is $1$, a finite, non-zero number. If the denominator still had an $x^1$ term (i.e., if $\alpha - 1 
eq 0$), the limit would be proportional to $x^2$, which goes to $0$ as $x 	o 0$.

But our limit is $1$. This means the $x^1$ term in the denominator must vanish. Therefore, $\alpha - 1 = 0$, which gives us **$\alpha = 1$**.

### The Final Cancellation

With $\alpha = 1$, the $x^1$ term disappears. Now, our limit simplifies to:

$$\lim_{x 	o 0} \frac{\beta x^3}{\frac{x^3}{6}} = \frac{\beta}{1/6} = 6\beta$$

Since the limit is given as $1$, we have $6\beta = 1$, which means **$\beta = \frac{1}{6}$**.

We have successfully navigated the complexity. The final step is simple arithmetic:

$$6(\alpha + \beta) = 6\left(1 + \frac{1}{6}ight) = 6 + 1 = \mathbf{7}$$

You see? It was never about brute force. It was about understanding the balance of powers. Keep this intuition with you, and no limit will ever intimidate you again.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $α, β \in R$ be such that $lim_{x \to 0} \frac{x ^{2} s i n ( β x )}{α x - s i n x} = 1$ . Then $6 (α + β)$ equals .........

Enter Numerical Value:

Visualized Solution

Identifying the Indeterminate Form

The Power of Series Expansion

Substituting the Expansions

Simplifying the Numerator

Simplifying the Denominator

The Condition for a Finite Limit

Solving for $α$

Solving for $β$

Final Calculation

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

Analyzing the Setup

The Gatekeeper

The Microscope

The Balancing Act

The Final Cancellation

Similar Questions

For $α, β, γ, \in R$ , if $lim_{x \to 0} \frac{x ^{2} s i n α x + ( γ - 1 ) e ^{x^{2}}}{s i n 2 x - β x} = 3$ , then $β + γ - α$ is equal to:

If $lim_{x \to 0} \frac{3 + α s i n x + β c o s x + l o g _{e} ( 1 - x )}{3 t a n ^{2} x} = \frac{1}{3}$ , then $2 α - β$ is equal to :

If $β = lim_{x \to 0} \frac{e ^{x^{3}} - ( 1 - x ^{3} ) ^{1/3} + ( ( 1 - x ^{2} ) ^{1/2} - 1 ) s i n x}{x s i n ^{2} x}$ , then the value of $6 β$ is ______.

The value of $lim_{x \to 0} ((sin x)^{1/ x} + (1/ x)^{s i n x})$ , where $x > 0$ is

Let $β = lim_{x \to 0} \frac{α x - ( e ^{3 x} - 1 )}{α x ( e ^{3 x} - 1 )}$ for some $α \in R$ . Then the value of $α + β$ is :

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 ?

Let $k \in R$ . If $lim_{x \to 0^{+}} (sin (sin k x) + cos x + x)^{2/ x} = e^{6}$ , then the value of $k$ is

If $lim_{x \to 0} \frac{α x e ^{x} - β l o g _{e} ( 1 + x ) + γ x ^{2} e ^{- x}}{x s i n ^{2} x} = 10$ , $α, β, γ \in R$ , then the value of $α + β + γ$ is

If $lim_{x \to 0} \frac{e ^{a x} - c o s ( b x ) - \frac{c x e ^{- c x}}{2}}{1 - c o s ( 2 x )} = 17$ , then $5 a^{2} + b^{2}$ is equal to

$lim_{x \to \frac{π}{4}} \frac{8 2 - ( c o s x + s i n x ) ^{7}}{2 - 2 s i n 2 x}$ is equal to

Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α,β∈R be such that limx→0​αx−sinxx2sin(βx)​=1. Then 6(α+β) equals .........