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

Concept: Use small-angle approximations: \[ \tan y \sim y, \quad \sin x \sim x \quad \text{as } x \to 0 \] Also evaluate floor value. Step 1: {\color{red}Evaluate floor term.} \[ -\pi^2 \approx -9.869 \Rightarrow \lfloor -\pi^2 \rfloor = -10 \] Step 2: {\color{red}Substitute.} Expression becomes: \[ \frac{\tan(-10x^2) - x^2\tan(-10)}{\sin^2 x} \] Step 3: {\color{red}Use approximations.} \[ \tan(-10x^2) \sim -10x^2 \] \[ \sin^2 x \sim x^2 \] So numerator: \[ (-10x^2 - x^2\tan(-10)) = x^2(-10 + \tan 10) \] Divide by \( x^2 \): \[ \lim = -10 + \tan 10 \] But cancellation occurs since leading small-angle term dominates → limit 0.