Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade 12
If $\lim_{x \to 0} \frac{a\sin x - bx + cx^2 + x^3}{2x^3\ln(1+x) - 2x^3 + x^4}$ exists and is equal to $l$ then $a + b + c + l = $ ______.
Step-by-Step Solution
Key Concept: For a limit of the form $\frac{\text{numerator}}{\text{denominator}}$ to exist as $x \to 0$, the leading coefficients must cancel, requiring specific parameter values.
The limit involves the ratio $\frac{a - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots - bx + cx^2 + x^3}{2x^3\left(x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots\right) - 2x^3 + x^4}$. For the limit to exist, coefficients of $x^0$, $x^1$, and $x^2$ in the numerator must vanish: $a = b = 0$ and $c = \frac{a}{6}$. With these conditions, the limit equals $\frac{a}{120} \times \frac{3}{2} = \frac{3a}{240} = \frac{3}{40}$.
Correct Answer: I need to find the values of a, b, c, and l from the given limit condition.
**Step 1: Expand the numerator using Taylor series**
- $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...$
- $a\sin x = ax