Practicing Success
$\underset{x→0}{\lim}\frac{\tan([-π^2]x^2)-\tan([-π^2])x^2}{\sin^2x}$ is equal to, ([x] represents greatest integer function) |
10 + tan 10 tan 10 −10 10 − tan 10 none of these |
tan 10 −10 |
$\underset{x→0}{\lim}\frac{\tan([-π^2]x^2)-\tan([-π^2]).x^2}{\sin^2x}$ Since $9<π^2<10$ $∴-10<-π^2<-9;\,∴\,[-π^2]=-10$ $=\underset{x→0}{\lim}\frac{\tan(-10x^2)-\tan(-10).x^2}{\sin^2x}$ $=\underset{x→0}{\lim}\frac{-\tan(10x^2)+\tan(10).x^2}{x^2.\frac{\sin^2x}{x^2}}=\underset{x→0}{\lim}\begin{bmatrix}\tan 10-\begin{pmatrix}\frac{\tan(10x^2)}{10x^2}\end{pmatrix}.10\end{bmatrix}=\tan 10-10$ |