Question

Q34. Let [x] denote the greatest integer less than or equal to x. Then: $\displaystyle{\sf{\lim_{x\to\infty}\dfrac{\tan(\pi\sin^2x)+(|x|-sin(x[x]))^2}{x^2}}}$

(1) equals π
(2) equals 0
(3) equals π + 1
(4) does not exist​

Answer 1

Appropriate Question

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

$\rm :\longmapsto\:\displaystyle{\sf{\lim_{x\to \: 0}\dfrac{\tan(\pi\sin^2x)+(|x|-sin(x[x]))^2}{x^2}}}$

(1) equals π

(2) equals 0

(3) equals π + 1

(4) does not exist

$\large\underline{\sf{Solution-}}$

Given statement is

$\rm :\longmapsto\:\displaystyle{\sf{\lim_{x\to \: 0}\dfrac{\tan(\pi\sin^2x)+(|x|-sin(x[x]))^2}{x^2}}}$

Consider RHL

$\rm :\longmapsto\:\displaystyle{\sf{\lim_{x\to \: 0^ + }\dfrac{\tan(\pi\sin^2x)+(|x|-sin(x[x]))^2}{x^2}}}$

$\red{\rm :\longmapsto\:Put \: x = 0 + h, \: as \: x \to \: 0, \: h \to \: 0}$

We know,

$\rm :\longmapsto\:[x] = 0 \: as \: x \: \to \: 0^ +$

$\rm \: = \:\displaystyle{\sf{\lim_{h\to \: 0}\dfrac{\tan(\pi\sin^2h)+(|h|-sin(0))^2}{h^2}}}$

$\rm \: = \:\displaystyle{\sf{\lim_{h\to \: 0}\dfrac{\tan(\pi\sin^2h)+(|h|-0)^2}{h^2}}}$

$\rm \: = \:\displaystyle{\sf{\lim_{h\to \: 0}\dfrac{\tan(\pi\sin^2h)+(|h|)^2}{h^2}}}$

$\rm \: = \:\displaystyle{\sf{\lim_{h\to \: 0}\dfrac{\tan(\pi\sin^2h)+h^2}{h^2}}}$

$\rm \: = \:\displaystyle{\sf{\lim_{h\to \: 0}\dfrac{\tan(\pi\sin^2h)}{h^2}}} + 1$

$\rm \: = \:\displaystyle{\sf{\lim_{h\to \: 0}\dfrac{\tan(\pi\sin^2h)}{\pi\sin^2h}}} \times \dfrac{\pi\sin^2h}{ {h}^{2} } + 1$

We know,

$\boxed{ \bf{\displaystyle{\sf{\lim_{x\to \: 0} \: \frac{tanx}{x} = 1}}}}$

$\boxed{ \bf{\displaystyle{\sf{\lim_{x\to \: 0} \: \frac{sinx}{x} = 1}}}}$

$\rm \: \: = \:\pi + 1$

So,

$\rm :\longmapsto\:\displaystyle{\sf{\lim_{x\to \: 0^ + }\dfrac{\tan(\pi\sin^2x)+(|x|-sin(x[x]))^2}{x^2}}} = \pi + 1$

Consider LHL

$\rm :\longmapsto\:\displaystyle{\sf{\lim_{x\to \: 0^ - }\dfrac{\tan(\pi\sin^2x)+(|x|-sin(x[x]))^2}{x^2}}}$

$\red{\rm :\longmapsto\:Put \: x = 0 - h, \: as \: x \to \: 0, \: h \to \: 0}$

We know,

$\rm :\longmapsto\:[x] = - 1 \: as \: x \: \to \: 0^ -$

$\rm \: = \:\displaystyle{\sf{\lim_{h\to \: 0}\dfrac{\tan(\pi\sin^2( - h))+(| - h|-sin( - h( - 1)))^2}{( - h)^2}}}$

$\rm \: = \:\displaystyle{\sf{\lim_{h\to \: 0}\dfrac{\tan(\pi\sin^2h)+(h-sin(h))^2}{h^2}}}$

$\rm \: = \:\displaystyle{\sf{\lim_{h\to \: 0}\dfrac{\tan(\pi\sin^2h)+(h-sinh)^2}{h^2}}}$

$\rm \: = \:\displaystyle{\sf{\lim_{h\to \: 0}\dfrac{\tan(\pi\sin^2h)+ {h}^{2} (1- \dfrac{sinh}{h} )^2}{h^2}}}$

$\rm \: = \:\displaystyle{\sf{\lim_{h\to \: 0}\dfrac{\tan(\pi\sin^2h)}{h^2}}} + (1 - 1)^{2}$

$\rm \: = \:\displaystyle{\sf{\lim_{h\to \: 0}\dfrac{\tan(\pi\sin^2h)}{\pi\sin^2h}}} \times \dfrac{\pi\sin^2h}{ {h}^{2} } + 0$

$\rm \: \: = \:\pi$

$\rm :\longmapsto\:\displaystyle{\sf{\lim_{x\to \: 0^ - }\dfrac{\tan(\pi\sin^2x)+(|x|-sin(x[x]))^2}{x^2}}} = \pi$

Thus, we concluded that

$\bf :\longmapsto\:LHL \: \ne \: RHL$

Hence,

$\bf :\longmapsto\:Limit \: doesnot \: exist.$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{\boxed{\bf{ Option \: (4) \: is \: correct }}}$