$\underset{x→0}{\lim}\frac{\tan([-π^2]x^2)-\tan([-π^2])x^2}{\sin^2x}$ is equal to, ([x] represents greatest integer function)
Answer & explanation
Correct answer: option 2
$\underset{x→0}{\lim}\frac{\tan([-π^2]x^2)-\tan([-π^2]).x^2}{\sin^2x}$
Since $9<π^2<10$
$∴-10<-π^2<-9;\,∴\,[-π^2]=-10$
$=\underset{x→0}{\lim}\frac{\tan(-10x^2)-\tan(-10).x^2}{\sin^2x}$
$=\underset{x→0}{\lim}\frac{-\tan(10x^2)+\tan(10).x^2}{x^2.\frac{\sin^2x}{x^2}}=\underset{x→0}{\lim}\begin{bmatrix}\tan 10-\begin{pmatrix}\frac{\tan(10x^2)}{10x^2}\end{pmatrix}.10\end{bmatrix}=\tan 10-10$