A) equals 0
B) equals\[\pi \]
C) equals \[\pi +1\]
D) does not exist
Correct Answer: D
Solution :
Here, \[R.H.L.=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\tan (\pi si{{n}^{2}}x)+{{\left( |x|-\sin (x\left[ x \right]) \right)}^{2}}}{{{x}^{2}}}\] \[=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\tan (\pi si{{n}^{2}}x)+{{x}^{2}}}{{{x}^{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because x\to {{0}^{+}}\Rightarrow [x]=0)\] \[=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\tan (\pi si{{n}^{2}}x)}{(\pi si{{n}^{2}}x)}.\frac{\pi si{{n}^{2}}x}{{{x}^{2}}}+1=\pi +1\] Now,\[L.H.L.=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{\tan (\pi si{{n}^{2}}x)+{{(-x+sinx)}^{2}}}{{{x}^{2}}}\] \[(\because x\to {{0}^{-}}\Rightarrow [x]=-1)\] \[=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{\tan (\pi si{{n}^{2}}x)}{\pi {{\sin }^{2}}x}.\frac{\pi si{{n}^{2}}x}{{{s}^{2}}}+{{\left( -1+\frac{\sin x}{x} \right)}^{2}}=\pi \] \[\therefore \]\[R.H.L.\ne L.H.L.\] So, limit does not exist.
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