Let [x] denote the greatest integer less than or equal to x Then … - Vidyadip

MCQ

Let $[x]$ denote the greatest integer less than or equal to $x$ Then

$\mathop {\lim }\limits_{x \to 0} \,\frac{{\tan \,(\pi \,{{\sin }^2}\,x) + \,{{(\left| x \right|\, - \,\sin \,(x\,[x]))}^2}}}{{{x^2}}}$

✓
does not exist
B
equals $\,\,\pi $
C
equals $\,\,\pi \,+\,1$
D
equals $\,\,0$

✓

Answer

Correct option: A.

does not exist

a
$\mathop {Lt}\limits_{x \to 0} \frac{{\tan \left( {\pi {{\sin }^2}x} \right)}}{{\pi {{\sin }^2}x}}.\frac{{\pi {{\sin }^2}x}}{{{x^2}}} + {\left( {\frac{{\left| x \right| - \sin \left( {x\left[ x \right]} \right)}}{{\left| x \right|}}} \right)^2}$

$1\left( \pi \right).{\left( 1 \right)^2} + \mathop {Lt}\limits_{x \to 0} {\left( {1 - \frac{{\sin x\left[ x \right]}}{{x\left[ x \right]}}.\frac{{x\left[ x \right]}}{{\left| x \right|}}} \right)^2}\,\,\,\,\,\,\,\,......\left( i \right)$

$\mathop {Lt}\limits_{x \to {0^ - }} \frac{{x\left[ x \right]}}{{\left| x \right|}} = \frac{x}{{ - x}}\left( { - 1} \right) = 1$

$\mathop {Lt}\limits_{x \to {0^ + }} \frac{{x\left[ x \right]}}{{\left| x \right|}} = \frac{{x\left( 0 \right)}}{x} = 0$

Put in equation $(i)$

$\therefore $ Limit does not exist.

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