JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Limits

Split the expression into $\frac{\tan (\pi {\sin }^{2}x)}{x^2}$ and ${\left(\frac{|x|-\sin (x[x])}{x}\right)}^2$. For the first, $\tan (\pi {\sin }^{2}x)\sim \pi {\sin }^{2}x\sim \pi x^2$, so it tends to $\pi$. The second depends on the side of approach through $[x]$: as $x\to 0^{-}$, $[x]=-1$ and $\frac{\sin (x[x])}{|x|}=\frac{\sin (-x)}{-x}=1$, giving $0$; as $x\to 0^{+}$, $[x]=0$ and the term is ${\left(1-0\right)}^2=1$. The left and right limits of the whole expression are $\pi$ and $\pi +1$, which disagree, so the limit does not exist.