For α, β, γ ∈ R, if lim x→0 (x^2 sin(αx) + (γ-1)e^(x^2)) / (sin(2x) - βx) = 3, then β + γ - α is equal to :

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q7 – JEE-MAIN 2025

← Back

Home Maths Limits and DerivativesQuestion #273

For

α, β, γ \in R

, if

lim_{x \to 0} \frac{x^{2} \sin (α x) + (γ - 1) e^{x^{2}}}{\sin (2 x) - β x} = 3

, then

β + γ - α

is equal to :

Clue 1 of 3: Indeterminate Form

The given limit is of the form $\frac{0}{0}$ as $x \to 0$ . For the limit to exist and be finite, the numerator must also approach zero.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Analyze the Indeterminate Form and Determine

γ

✦ Active

The given limit is $lim_{x \to 0} \frac{x^{2} \sin (α x) + (γ - 1) e^{x^{2}}}{\sin (2 x) - β x} = 3$ . As $x \to 0$ , the denominator $\sin (2 x) - β x \to 0 - 0 = 0$ . For the limit to be finite, the numerator must also approach 0. Evaluating the numerator at $x = 0$ : $0^{2} \sin (0) + (γ - 1) e^{0} = 0 + (γ - 1) (1) = γ - 1$ . Therefore, we must have $γ - 1 = 0$ , which implies $γ = 1$ .

Step 2: Apply Taylor Series Expansions and Determine

β

○ Expand

Substitute $γ = 1$ into the limit expression. The numerator becomes $x^{2} \sin (α x)$ . Now, expand the numerator and denominator using Taylor series around $x = 0$ :

\sin (α x) = α x - \frac{(α x)^{3}}{3!} + O (x^{5})

x^{2} \sin (α x) = x^{2} (α x - \frac{α^{3} x^{3}}{6} + O (x^{5})) = α x^{3} - \frac{α^{3} x^{5}}{6} + O (x^{7})

\sin (2 x) = 2 x - \frac{(2 x)^{3}}{3!} + O (x^{5}) = 2 x - \frac{8 x^{3}}{6} + O (x^{5}) = 2 x - \frac{4 x^{3}}{3} + O (x^{5})

Denominator = (2 x - \frac{4 x^{3}}{3} + O (x^{5})) - β x = (2 - β) x - \frac{4 x^{3}}{3} + O (x^{5})

The limit expression is now $lim_{x \to 0} \frac{α x^{3} - \frac{α^{3} x^{5}}{6} + O (x^{7})}{(2 - β) x - \frac{4 x^{3}}{3} + O (x^{5})}$ . For the limit to be finite and non-zero, the lowest power of $x$ in the denominator must match the lowest power in the numerator (which is $x^{3}$ ). This means the $x$ term in the denominator must be zero. Thus, $2 - β = 0$ , which implies $β = 2$ .

💡 Teacher's Secret Hint

Alternatively, L'Hopital's rule can be applied multiple times until the denominator is non-zero at $x = 0$ .

Step 3: Calculate

α

and the Final Expression○ Expand

With $γ = 1$ and $β = 2$ , the limit expression simplifies to:

lim_{x \to 0} \frac{α x^{3} - \frac{α^{3} x^{5}}{6} + O (x^{7})}{- \frac{4 x^{3}}{3} + O (x^{5})}

= lim_{x \to 0} \frac{α x^{3}}{- \frac{4 x^{3}}{3}} = \frac{α}{- \frac{4}{3}} = - \frac{3 α}{4}

We are given that this limit is equal to 3. So, $- \frac{3 α}{4} = 3 ⟹ - 3 α = 12 ⟹ α = - 4$ . Now, we need to find $β + γ - α$ :

β + γ - α = 2 + 1 - (- 4) = 3 + 4 = 7

💡 Teacher's Secret Hint

Ensure all signs are handled correctly when substituting values.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 7 – JEE-MAIN 2025

Clue 1 of 3: Indeterminate Form

🧠 Full Solution Path