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Limits: Let Following Correct

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

Question
Let L=lim⁡x→1sin⁡(6cos⁡−1x)1−x2L=\displaystyle\lim_{x\to 1}\dfrac{\sin(6\cos^{-1}x)}{\sqrt{1-x^{2}}} and M=lim⁡x→11−cos⁡(6cos⁡−1x)1−x2M=\displaystyle\lim_{x\to 1}\dfrac{1-\cos(6\cos^{-1}x)}{1-x^{2}}. Which of the following is/are correct?
AL+M=24L+M=24correct
BML=3\dfrac{M}{L}=3correct
CL−M=−12L-M=-12correct
DLM=108LM=108correct
Solution
Step 1: Substitute cos⁡−1x=θ\cos^{-1}x=\theta, so x=cos⁡θx=\cos\theta. As x→1x\to 1, θ→0\theta\to 0. Step 2: Compute LL:
L=lim⁡θ→0sin⁡6θ1−cos⁡2θ=lim⁡θ→0sin⁡6θ∣sin⁡θ∣=lim⁡θ→0sin⁡6θsin⁡θL=\lim_{\theta\to 0}\dfrac{\sin 6\theta}{\sqrt{1-\cos^{2}\theta}}=\lim_{\theta\to 0}\dfrac{\sin 6\theta}{|\sin\theta|}=\lim_{\theta\to 0}\dfrac{\sin 6\theta}{\sin\theta}
 Step 3: Use sin⁡6θsin⁡θ=sin⁡6θ6θ⋅θsin⁡θ⋅6\dfrac{\sin 6\theta}{\sin\theta}=\dfrac{\sin 6\theta}{6\theta}\cdot\dfrac{\theta}{\sin\theta}\cdot 6:
L=1⋅1⋅6=6L=1\cdot 1\cdot 6=6
 Step 4: Compute MM:
M=lim⁡θ→01−cos⁡6θ1−cos⁡2θ=lim⁡θ→01−cos⁡6θsin⁡2θM=\lim_{\theta\to 0}\dfrac{1-\cos 6\theta}{1-\cos^{2}\theta}=\lim_{\theta\to 0}\dfrac{1-\cos 6\theta}{\sin^{2}\theta}
 Step 5: Use 1−cos⁡6θ=2sin⁡23θ1-\cos 6\theta=2\sin^{2}3\theta:
M=lim⁡θ→02sin⁡23θsin⁡2θ=2lim⁡θ→0(sin⁡3θsin⁡θ)2=2⋅9=18M=\lim_{\theta\to 0}\dfrac{2\sin^{2}3\theta}{\sin^{2}\theta}=2\lim_{\theta\to 0}\left(\dfrac{\sin 3\theta}{\sin\theta}\right)^{2}=2\cdot 9=18
 Step 6: So L=6L=6 and M=18M=18. Check options: (1) L+M=6+18=24L+M=6+18=24. True. (2) ML=186=3\dfrac{M}{L}=\dfrac{18}{6}=3. True. (3) L−M=6−18=−12L-M=6-18=-12. True. (4) LM=6⋅18=108LM=6\cdot 18=108. True. Correct answers: (1), (2), (3) and (4)
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