Let $[\mathrm{x}]$ denote the greatest integer less than or equal to $x$. Then :-
$\lim _{x \rightarrow 0} \frac{\tan \left(\pi \sin ^{2} x\right)+(|x|-\sin (x[x]))^{2}}{x^{2}}$
Correct Option: , 4
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.
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