Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade 12
If $\lim_{x \to \infty} 4x\left(\frac{\pi}{4} - \tan^{-1}\frac{x+1}{x+2}\right) = y^2 + 4y + 5$, then the product of all possible value of $y$ is ______.
Step-by-Step Solution
Key Concept: Standard limit $\lim_{u \to 0} \frac{\tan^{-1}(u)}{u} = 1$ simplifies inverse trigonometric limits.
Given $\lim_{x \to a} 4x \cdot \frac{\tan^{-1}\left(\frac{1}{2x+3}\right)}{\frac{1}{2x+3}} \times \frac{1}{2x+3} = 2$, we use $\lim_{u \to 0} \frac{\tan^{-1}(u)}{u} = 1$. This simplifies to $4a \times 1 \times \frac{1}{2a+3} = 2$, yielding $\frac{4a}{2a+3} = 2$. Solving: $4a = 4a + 6$, which gives $y^2 + 4y + 5 = 2$, so $y = -1$ or $y = -3$.
Correct Answer: Looking at this problem, I need to evaluate the limit and solve for y.
**Step 1: Simplify the inverse tangent expression**
$$\frac{\pi}{4} - \tan^{-1}\frac{x+1}{x+2}$$
Using the identity $\frac{\pi}{4} = \tan^{-