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
- A \(0\)
- B \(\tan 10-10\)
- C \(\tan 9-9\)
- D 1
Answer & Solution
Correct Answer
(B) \(\tan 10-10\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \lim _{x \rightarrow 0} \frac{-\tan 10 x^2+x^2 \tan 10}{\sin ^2 x} \quad \text { (Applying series) } \\ & \lim _{x \rightarrow 0} \frac{-\left(10 x^2+\frac{10^3 x^6}{3}\right)+x^2 \tan 10}{x^2\left(1-\frac{x^2}{6}\right)^2}=\tan 10-10\end{aligned}\)
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