Question

Evaluate \[ \lim_{x \to 0} \frac{\tan\!\left(\lfloor -\pi^2 \rfloor x^2\right) - x^2 \tan\!\left(\lfloor -\pi^2 \rfloor\right)}{\sin^2 x} \]

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

Hint:

For limits with floor functions: \begin{itemize} \item Evaluate floor first. \item Then apply small-angle expansions. \end{itemize}

Answer 1

The Correct Option is A