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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
lim⁡x→0[∣sin⁡−1x∣x]\displaystyle\lim_{x\to 0}\left[\dfrac{|\sin^{-1}x|}{x}\right], where [⋅][\cdot] denotes greatest integer function, is
ALeft hand limit is −2-2correct
BLeft hand limit is −1-1
CRight hand limit is 11correct
DLimit exists and both are equal to 11
Solution
Step 1: For small x>0x>0: sin⁡−1x>x>0\sin^{-1}x>x>0, so sin⁡−1xx>1\dfrac{\sin^{-1}x}{x}>1. More precisely, the Taylor series gives sin⁡−1x=x+x36+…\sin^{-1}x=x+\dfrac{x^{3}}{6}+\ldots, so sin⁡−1xx=1+x26+…\dfrac{\sin^{-1}x}{x}=1+\dfrac{x^{2}}{6}+\ldots, which is just slightly greater than 11. Step 2: For x>0x>0: ∣sin⁡−1x∣=sin⁡−1x|\sin^{-1}x|=\sin^{-1}x, so ∣sin⁡−1x∣x>1\dfrac{|\sin^{-1}x|}{x}>1 but very close to 11.
RHL: [∣sin⁡−1x∣x]=1\text{RHL: }\left[\dfrac{|\sin^{-1}x|}{x}\right]=1
 Option (3) is correct. Step 3: For x<0x<0: sin⁡−1x<0\sin^{-1}x<0, so ∣sin⁡−1x∣=−sin⁡−1x|\sin^{-1}x|=-\sin^{-1}x. And −sin⁡−1x>0-\sin^{-1}x>0 while x<0x<0, so
∣sin⁡−1x∣x=−sin⁡−1xx<0\dfrac{|\sin^{-1}x|}{x}=\dfrac{-\sin^{-1}x}{x}<0
 Step 4: Let h=−x>0h=-x>0, then sin⁡−1(−h)=−sin⁡−1h\sin^{-1}(-h)=-\sin^{-1}h, so ∣sin⁡−1(−h)∣=sin⁡−1h|\sin^{-1}(-h)|=\sin^{-1}h and the ratio becomes sin⁡−1h−h\dfrac{\sin^{-1}h}{-h}, which is slightly less than −1-1 (since sin⁡−1h>h\sin^{-1}h>h slightly). Step 5: So for x<0x<0: ∣sin⁡−1x∣x∈(−1−ϵ,−1)\dfrac{|\sin^{-1}x|}{x}\in(-1-\epsilon,-1), meaning between −2-2 and −1-1. Therefore
LHL: [∣sin⁡−1x∣x]=−2\text{LHL: }\left[\dfrac{|\sin^{-1}x|}{x}\right]=-2
 Option (1) is correct. Step 6: Since LHL =−2≠1==-2\neq 1= RHL, limit does not exist. Option (D) is wrong. Correct answers: (1) and (3)
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