The value of $\lim_{x \to 0} \frac{1-\cos(\tan x)}{x^2 \sin(2\sin(x)\cdot \cos(\frac{1}{x}))}$ is equal to
Step-by-Step Solution
Key Concept: Using Taylor series expansions and understanding the behavior of bounded oscillatory functions like $\cos(\frac{1}{x})$ in limit problems.
Using Taylor expansions near $x = 0$: $\tan x \approx x + \frac{x^3}{3} + \ldots$ and $\cos(\tan x) \approx 1 - \frac{x^2}{2} + \ldots$, so the numerator is $1 - \cos(\tan x) \approx \frac{x^2}{2}$. For the denominator, $\sin(2\sin(x)\cdot\cos(\frac{1}{x}))$ is bounded by $|\sin(x)|$ which is $O(x)$, making the denominator $O(x^3)$ or higher order. Applying L'Hôpital's rule or careful expansion yields the limit as $\frac{1}{4}$.
Correct Answer: 1