Question

The value of $\underset{x\to 0}{lim}\frac{\tan \left(\pi {\sin }^{2}x\right)+(|x|-\sin (x\left[x\right]){)}^2}{x^2}=$
(where $[.]$ is greatest integer function)

• A. $\pi$
• B. $0$
• C. $\pi +1$
• D. Does not exist.

Answer 1

The correct option is D Does not exist.

Given : $\underset{x\to 0}{lim}\frac{\tan \left(\pi {\sin }^{2}x\right)+(|x|-\sin (x\left[x\right]){)}^2}{x^2}$

$L.H.L.=\underset{x\to 0^{-}}{lim}\frac{\tan \left(\pi {\sin }^{2}x\right)+(-x-\sin x(-1){)}^2}{x^2}=\underset{x\to 0^{-}}{lim}[\frac{\tan \left(\pi {\sin }^{2}x\right)}{\pi {\sin }^{2}x}\times \frac{\pi {\sin }^{2}x}{x^2}]+{[-1+\frac{\sin x}{x}]}^2=\pi +(-1+1{)}^2=\pi$

$R.H.L.=\underset{x\to 0^{+}}{lim}\frac{\tan \left(\pi {\sin }^{2}x\right)+(x-0{)}^2}{x^2}=\underset{x\to 0^{+}}{lim}\frac{\tan \left(\pi {\sin }^{2}x\right)}{\pi {\sin }^{2}x}\times \frac{\pi {\sin }^{2}x}{x^2}+{\left(1\right)}^2=\pi +(1{)}^2=\pi +1$

$\therefore L.H.L.\ne R.H.L.$
So, limit does not exist.