Let [ x] denote the greatest integer less than or equal to X. Then : lim _x → 0 tan (π sin ^2 x )+(| x|-sin (x[x]))^2 x^2

RHL is, $\displaystyle \lim _{x \rightarrow 0^{+}} \frac{\tan \left(\pi \sin ^{2} x\right)+(x-0)^{2}}{x^{2}}$x→0+lim​x2tan(πsin2x)+(x−0)2​ $\displaystyle =\lim _{x \rightarrow 0^{+}}\left(\frac{\tan \left(\pi \sin ^{2} x\right.}{x^{2}}+1\right)=1+\pi$=x→0+lim​(x2tan(πsin2x​+1)=1+π And LHL is, $\displaystyle \lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi \sin ^{2} x\right)+(-x+\sin x)^{2}}{x^{2}}$x→0−lim​x2tan(πsin2x)+(−x+sinx)2​ $\displaystyle =\lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi \sin ^{2} x\right)+x^{2}+\sin ^{2} x-2 x \sin x}{x^{2}}$=x→0−lim​x2tan(πsin2x)+x2+sin2x−2xsinx​ $=\pi+1+1-2=\pi$=π+1+1−2=π Since, LHL $\neq$= RHL Hence, limit does not exist.