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Limits Continuity and Differentiability question

2019 · 11 Jan · Shift 1 · Q22
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  5. /2019 · 11 Jan · Shift 1 · Q22

Limits Continuity and Differentiability question

2019 · 11 Jan · Shift 1 · Q22

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let [x] denote the greatest integer less than or equal to x. Then lim⁡x→0tan⁡(πsin⁡2x)+(∣x∣−sin⁡(x[x]))2x2\mathop {\lim }\limits_{x \to 0} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}x→0lim​x2tan(πsin2x)+(∣x∣−sin(x[x]))2​
  1. A
    equals π\piπ + 1
  2. B
    equals 0
  3. C
    does not exist
  4. D
    equals π\piπ
View written solutionFree

Correct answer: C

  1. We need to evaluate L=lim⁡x→0tan⁡(πsin⁡2x)+(∣x∣−sin⁡(x[x]))2x2.L=\lim_{x\to 0}\frac{\tan\big(\pi\sin^2 x\big)+\Big(|x|-\sin(x[x])\Big)^2}{x^2}.L=limx→0​x2tan(πsin2x)+(∣x∣−sin(x[x]))2​.

  2. The key issue is the floor function [x][x][x] near x=0x=0x=0.

  • If x→0+x\to 0^+x→0+, then [x]=0[x]=0[x]=0.
  • If x→0−x\to 0^-x→0−, then [x]=−1[x]=-1[x]=−1.

So we compute right-hand and left-hand limits separately.

3. Right-hand limit: x→0+x\to 0^+x→0+

For x>0x>0x>0 small, [x]=0[x]=0[x]=0. Hence x[x]=x⋅0=0,sin⁡(x[x])=sin⁡0=0.x[x]=x\cdot 0=0, \qquad \sin(x[x])=\sin 0=0.x[x]=x⋅0=0,sin(x[x])=sin0=0. Therefore (∣x∣−sin⁡(x[x]))2=(x−0)2=x2.\Big(|x|-\sin(x[x])\Big)^2=(x-0)^2=x^2.(∣x∣−sin(x[x]))2=(x−0)2=x2. So the expression becomes

=\frac{\tan(\pi\sin^2 x)}{x^2}+1.$$ Now as $x\to 0$, $$\sin x\sim x \implies \sin^2 x\sim x^2,$$ so $$\pi\sin^2 x\sim \pi x^2.$$ Also, for small $u$, $\tan u\sim u$. Thus $$\tan(\pi\sin^2 x)\sim \pi\sin^2 x\sim \pi x^2.$$ Hence $$\lim_{x\to 0^+}\frac{\tan(\pi\sin^2 x)}{x^2}=\pi.$$ Therefore $$\lim_{x\to 0^+}\frac{\tan(\pi\sin^2 x)+\big(|x|-\sin(x[x])\big)^2}{x^2}=\pi+1.$$ --- ### 4. Left-hand limit: $x\to 0^-$ For $x<0$ small, $[x]=-1$. Hence $$x[x]=x(-1)=-x,$$ and therefore $$\sin(x[x])=\sin(-x).$$ Since $x<0$, we have $|x|=-x$. Thus $$|x|-\sin(x[x])=-x-\sin(-x).$$ Using $\sin(-x)=-\sin x$, $$-x-\sin(-x)=-x+\sin x=\sin x-x.$$ So $$\Big(|x|-\sin(x[x])\Big)^2=(\sin x-x)^2.$$ Hence the expression becomes $$\frac{\tan(\pi\sin^2 x)+(\sin x-x)^2}{x^2}.

Now, tan⁡(πsin⁡2x)∼πx2,\tan(\pi\sin^2 x)\sim \pi x^2,tan(πsin2x)∼πx2, so tan⁡(πsin⁡2x)x2→π.\frac{\tan(\pi\sin^2 x)}{x^2}\to \pi.x2tan(πsin2x)​→π. Also, sin⁡x−x∼−x36,\sin x-x\sim -\frac{x^3}{6},sinx−x∼−6x3​, therefore (sin⁡x−x)2∼x636,(\sin x-x)^2\sim \frac{x^6}{36},(sinx−x)2∼36x6​, so (sin⁡x−x)2x2∼x436→0.\frac{(\sin x-x)^2}{x^2}\sim \frac{x^4}{36}\to 0.x2(sinx−x)2​∼36x4​→0. Thus lim⁡x→0−tan⁡(πsin⁡2x)+(∣x∣−sin⁡(x[x]))2x2=π.\lim_{x\to 0^-}\frac{\tan(\pi\sin^2 x)+\big(|x|-\sin(x[x])\big)^2}{x^2}=\pi.limx→0−​x2tan(πsin2x)+(∣x∣−sin(x[x]))2​=π.

5. Compare one-sided limits

  • Right-hand limit =π+1=\pi+1=π+1
  • Left-hand limit =π=\pi=π

Since these are unequal, the two-sided limit does not exist.

6. Check options

  • A: π+1\pi+1π+1 — false (only right-hand limit)
  • B: 000 — false
  • C: does not exist — true
  • D: π\piπ — false (only left-hand limit)

Therefore, the correct answer is C.\boxed{\text{C}}.C​.

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