Question

Let $\left[x\right]$ denote the greatest integer less than or equal to $x$. Then :
$\underset{x\to 0}{lim}\frac{\tan \left(\pi {\sin }^{2}x\right)+(|x|-\sin (x\left[x\right]){)}^2}{x^2}$:

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

Answer 1

The correct option is D does not exist

$\text{LHL}=\underset{x\to 0^{-}}{lim}\frac{\tan \left(\pi {\sin }^{2}x\right)+(-x-\sin x(-1){)}^2}{x^2}$

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

$=1\times \pi +(-1+1{)}^2$
$=\pi$


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

$=1\times \pi +(1-0{)}^2$
$=\pi +1$

$\therefore \text{LHL}\ne \text{RHL}$
$\Rightarrow$ Limit does not exist.