Let the function
Question:

Let $[\mathrm{x}]$ denote the greatest integer less than or equal to $x$. Then :-

$\lim _{x \rightarrow 0} \frac{\tan \left(\pi \sin ^{2} x\right)+(|x|-\sin (x[x]))^{2}}{x^{2}}$

1. equals $\pi$

2. equals 0

3. equals $\pi+1$

4. does not exist

Correct Option: , 4

Solution:

R.H.L. $=\lim _{x \rightarrow 0^{+}} \frac{\tan \left(\pi \sin ^{2} x\right)+(|x|-\sin (x[x]))^{2}}{x^{2}}$

$\left(\right.$ as $\left.x \rightarrow 0^{+} \Rightarrow[x]=0\right)$

$=\lim _{x \rightarrow 0^{+}} \frac{\tan \left(\pi \sin ^{2} x\right)+x^{2}}{x^{2}}$

$=\lim _{x \rightarrow 0^{+}} \frac{\tan \left(\pi \sin ^{2} x\right)}{\left(\pi \sin ^{2} x\right)}+1=\pi+1$

L.H.L. $=\lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi \sin ^{2} x\right)+(-x+\sin x)^{2}}{x^{2}}$

(as $\left.x \rightarrow 0^{-} \Rightarrow[x]=-1\right)$

$\lim _{x \rightarrow 0+} \frac{\tan \left(\pi \sin ^{2} x\right)}{\pi \sin ^{2} x} \cdot \frac{\pi \sin ^{2} x}{x^{2}}+\left(-1+\frac{\sin x}{x}\right)^{2} \Rightarrow \pi$

R.H.L. $\neq$ L.H.L.