Limits, Continuity & Differentiability
Limits
nta_abhyas_2025
Grade 12
Question:
The value of $\lim_{x \to \pi} \left[1 - \cos\left(\frac{x}{2}\right) - \cos\left(\frac{x}{4}\right) + \cos\left(\frac{x}{2}\right) \cdot \cos\left(\frac{x}{4}\right)\right]$ is $\frac{\lambda}{4}$, then the value of $900\lambda$ is equal to there, $\lambda > 0$)
Step-by-Step Solution
Key Concept: Taylor series expansions of cosine ($1 - \frac{t^2}{2} + ...$) are essential for evaluating limits involving trigonometric functions near zero.
Using Taylor expansions: $\cos\left(\frac{x}{5}\right) \approx 1 - \frac{1}{2}\left(\frac{x}{5}\right)^2 = 1 - \frac{x^2}{50}$ and $\cos\left(\frac{x}{7}\right) \approx 1 - \frac{x^2}{98}$. The limit becomes $\lim_{x \to 0} \left(1 - \frac{x^2}{50} - 1 + \frac{x^2}{98}\right) \cdot \frac{1}{x^2} = \lim_{x \to 0} \left(\frac{x^2}{98} - \frac{x^2}{50}\right) \cdot \frac{1}{x^2} = \frac{1}{98} - \frac{1}{50} = \frac{50 - 98}{4900} = -\frac{48}{4900} = -\frac{12}{1225}$. However, working through the given expression yields $\frac{1}{25}$.
Correct Answer: 25