Question

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

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

$R.H.L.$
$=\underset{x\to 0^{+}}{lim}\frac{\tan \left(\pi {\sin }^{2}x\right)+(|x|-\sin (x\left[x\right]){)}^2}{x^2}$
$=\underset{h\to 0}{lim}\frac{\tan \left(\pi {\sin }^2\right(0+h\left)\right)+(|0+h|-\sin ((0+h)[0+h]){)}^2}{(0+h{)}^2}$
$=\underset{h\to 0}{lim}\frac{\tan \left(\pi {\sin }^{2}h\right)+(h-0{)}^2}{h^2}$
$=\underset{h\to 0}{lim}\frac{\tan \left(\pi {\sin }^{2}h\right)}{\pi {\sin }^{2}h}\times \frac{\pi {\sin }^{2}h}{h^2}+{\left(1\right)}^2$
$=\pi +(1{)}^2$
$=\pi +1$
$\therefore L.H.L.\ne R.H.L.$
So, limit does not exist.