Question

Let [$x$ ] denote the greatest integer less than or equal to $x$. Then :-
$\displaystyle\lim_{x\to0} \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\to0^{+}} \frac{\tan\left(\pi\sin^{2} x\right)+\left(\left|x\right|-\sin\left(x\left[x\right]\right)\right)^{2}}{x^{2}} $
(as $x \to 0^+$ $\Rightarrow $ [x] = 0)
$= \lim_{x\to0^{+}} \frac{\tan\left(\pi\sin^{2} x\right)+x^{2}}{x^{2}} $
$ =\lim_{x\to0^{+}} \frac{\tan\left(\pi\sin^{2}x\right)}{\left(\pi\sin^{2}x\right)} + 1 = \pi+1 $
$ L.H.L = \lim_{x\to0^{-}} \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\to0^{+} } \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.$