Question

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

• A. equals $\pi$
• B. equals $0$
• C. equals $\pi +1$
• D. does not exist

Answer 1

Answer: (D)

$R.H.L={\lim }_{x\to 0^{+}}\frac{\tan \left(\pi {\sin }^{2}x\right)+{\left(|x|-\sin \left(x\left[x\right]\right)\right)}^2}{x^2}$
(as $x\to 0^{+}$ $\Rightarrow$ [x] = 0)
$={\lim }_{x\to 0^{+}}\frac{\tan \left(\pi {\sin }^{2}x\right)+x^2}{x^2}$
$={\lim }_{x\to 0^{+}}\frac{\tan \left(\pi {\sin }^{2}x\right)}{\left(\pi {\sin }^{2}x\right)}+1=\pi +1$
$L.H.L={\lim }_{x\to 0^{-}}\frac{\tan \left(\pi {\sin }^{2}x\right)+{\left(-x+\sin x\right)}^2}{x^2}$
(as x $\to$ 0- $\Rightarrow$ [x] = -1)
${\lim }_{x\to 0^{+}}\frac{\tan \left(\pi {\sin }^{2}x\right)}{\pi {\sin }^{2}x}.\frac{\pi {\sin }^{2}x}{x^2}+{\left(-1+\frac{\sin x}{x}\right)}^2\Rightarrow \pi$
$R.H.L.\ne L.H.L.$