Question

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

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

Answer 1

Answer: (D)

does not exist

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

L.H.L. $=\underset{x\to 0^{-}}{lim}\frac{tan\left(\pi si{n}^{2}x\right)+(-x+sinx{)}^2}{x^2}$
(as $x\to 0-\Rightarrow \left[x\right]=-1$)
$\underset{x\to 0+}{lim}\frac{tan\left(\pi si{n}^{2}x\right)}{\pi si{n}^{2}x}.\frac{\pi si{n}^{2}x}{x^2}+{(-1+\frac{sinx}{x})}^2\Rightarrow \pi$

$R.H.L\ne L.H.L.$