Mathematics - Limits Question with Solution | TestHub

RHL is, ${\lim }_{x\to 0^{+}}\frac{\tan \left(\pi {\sin }^{2}x\right)+(x-0{)}^2}{x^2}$ $={\lim }_{x\to 0^{+}}\left(\frac{\tan (\pi {\sin }^{2}x}{x^2}+1\right)=1+\pi$ And LHL is, ${\lim }_{x\to 0^{-}}\frac{\tan \left(\pi {\sin }^{2}x\right)+(-x+\sin x{)}^2}{x^2}$ $={\lim }_{x\to 0^{-}}\frac{\tan \left(\pi {\sin }^{2}x\right)+x^2+{\sin }^{2}x-2x\sin x}{x^2}$ $=\pi +1+1-2=\pi$ Since, LHL $\ne$ RHL Hence, limit does not exist.