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WBJEE · Maths · Limits

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

  1. A \(0\)
  2. B \(\tan 10-10\)
  3. C \(\tan 9-9\)
  4. D 1
Verified Solution

Answer & Solution

Correct Answer

(B) \(\tan 10-10\)

Step-by-step Solution

Detailed explanation

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