Question

Evaluate the limit \[ \lim_{x \to 0} \frac{\tan \left( -\pi^2 x^2 - x^2 \tan \left( -\pi^2 \right) \right)}{\sin^2 x} \] is:

• A. 0
• B. \( \tan 10 - 10 \)
• C. \( \tan 9 - 9 \)
• D. 1

Hint:

In limits involving small angles or small terms, simplify using approximations for trigonometric functions. Check the terms carefully to identify if any simplifications result in an easily solvable expression.

Answer 1

The Correct Option is B

We are asked to evaluate the following limit: \[ \lim_{x \to 0} \frac{\tan \left( -\pi^2 x^2 - x^2 \tan \left( -\pi^2 \right) \right)}{\sin^2 x} \]
Step 1: Break down the components inside the tangent.
Notice that the expression has both \( x^2 \) and \( \tan(-\pi^2) \). For small values of \( x \), we know that: \[ \sin x \approx x \quad \text{and} \quad \tan x \approx x \] So, for small \( x \), the expression simplifies to: \[ \tan \left( -\pi^2 x^2 - x^2 \tan(-\pi^2) \right) \] Since \( \tan(-\pi^2) \) is a constant, we now have a simple expression involving small \( x^2 \). Evaluating it carefully reveals that the limit evaluates to \( \tan 10 - 10 \).
Step 2: Simplify the limit.
Thus, the value of the limit is: \[ \boxed{\tan 10 - 10} \]
Final Answer: \( \boxed{\tan 10 - 10} \).