Q81 / 155 · WBJEE 2025 · Mathematics

$\lim _{x \rightarrow 0} \frac{\tan \left(\left[-\pi^2\right] x^2\right)-x^2 \tan \left(\left[-\pi^2\right]\right)}{\sin ^2 x}$ equals

  • A
    $0$
  • Correct answer
    B
    $\tan 10-10$
  • C
    $\tan 9-9$
  • D
    1

Solution

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{-\tan 10 x^2+x^2 \tan 10}{\sin ^2 x} \quad \text { (Applying series) } \\ & \lim _{x \rightarrow 0} \frac{-\left(10 x^2+\frac{10^3 x^6}{3}\right)+x^2 \tan 10}{x^2\left(1-\frac{x^2}{6}\right)^2}=\tan 10-10\end{aligned}$

Hence, the correct answer is option: 'B'.

Relevant Questions