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Limits — JEE Maths practice question

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

Question
If α,β\alpha,\beta are the roots of ax2+bx+c=0ax^{2}+bx+c=0.
L1=lim⁡x→α1−cos⁡(ax2+bx+c)(x−α)2,L2=lim⁡x→β1−cos⁡(ax2+bx+c)(x−β)2L_{1}=\lim_{x\to\alpha}\dfrac{1-\cos(ax^{2}+bx+c)}{(x-\alpha)^{2}},\quad L_{2}=\lim_{x\to\beta}\dfrac{1-\cos(ax^{2}+bx+c)}{(x-\beta)^{2}}
L3=lim⁡(x−α)(x−β)→01−cos⁡(ax2+bx+c)(x−α)2(x−β)2. ThenL_{3}=\lim_{(x-\alpha)(x-\beta)\to 0}\dfrac{1-\cos(ax^{2}+bx+c)}{(x-\alpha)^{2}(x-\beta)^{2}}.\text{ Then}
AL1=L2L_{1}=L_{2}correct
BL1=L3(α−β)2L_{1}=L_{3}(\alpha-\beta)^{2}correct
CL2=L3(α−β)2L_{2}=L_{3}(\alpha-\beta)^{2}correct
DL3=a22L_{3}=\dfrac{a^{2}}{2}correct
Solution
Step 1: Since α,β\alpha,\beta are roots: ax2+bx+c=a(x−α)(x−β)ax^{2}+bx+c=a(x-\alpha)(x-\beta). Step 2: Use the identity 1−cos⁡θ=2sin⁡2θ21-\cos\theta=2\sin^{2}\dfrac{\theta}{2}, so as θ→0\theta\to 0, 1−cos⁡θ∼θ221-\cos\theta\sim\dfrac{\theta^{2}}{2}. Step 3: Compute L1L_{1}. As x→αx\to\alpha, ax2+bx+c=a(x−α)(x−β)→0ax^{2}+bx+c=a(x-\alpha)(x-\beta)\to 0:
L1=lim⁡x→α[a(x−α)(x−β)]2/2(x−α)2=a2(α−β)22L_{1}=\lim_{x\to\alpha}\dfrac{[a(x-\alpha)(x-\beta)]^{2}/2}{(x-\alpha)^{2}}=\dfrac{a^{2}(\alpha-\beta)^{2}}{2}
 Hmm, the book uses substitution of the simpler form. Let us recompute carefully:
L1=lim⁡x→α1−cos⁡a(x−α)(x−β)[a(x−α)(x−β)]2⋅[a(x−α)(x−β)]2(x−α)2L_{1}=\lim_{x\to\alpha}\dfrac{1-\cos a(x-\alpha)(x-\beta)}{[a(x-\alpha)(x-\beta)]^{2}}\cdot\dfrac{[a(x-\alpha)(x-\beta)]^{2}}{(x-\alpha)^{2}}
 The first factor →12\to\dfrac{1}{2} and the second →a2(α−β)2\to a^{2}(\alpha-\beta)^{2}. So
L1=a2(α−β)22L_{1}=\dfrac{a^{2}(\alpha-\beta)^{2}}{2}
 Step 4: By symmetry (swap α↔β\alpha\leftrightarrow\beta):
L2=a2(β−α)22=a2(α−β)22L_{2}=\dfrac{a^{2}(\beta-\alpha)^{2}}{2}=\dfrac{a^{2}(\alpha-\beta)^{2}}{2}
 So L1=L2L_{1}=L_{2}. Step 5: Compute L3L_{3}:
L3=lim⁡1−cos⁡a(x−α)(x−β)(x−α)2(x−β)2=a22L_{3}=\lim_{}\dfrac{1-\cos a(x-\alpha)(x-\beta)}{(x-\alpha)^{2}(x-\beta)^{2}}=\dfrac{a^{2}}{2}
 Step 6: Check options: (1) L1=L2L_{1}=L_{2}: True. (2) L1=L3(α−β)2=a22(α−β)2L_{1}=L_{3}(\alpha-\beta)^{2}=\dfrac{a^{2}}{2}(\alpha-\beta)^{2}: True. (3) L2=L3(α−β)2=a22(α−β)2L_{2}=L_{3}(\alpha-\beta)^{2}=\dfrac{a^{2}}{2}(\alpha-\beta)^{2}: True. (4) L3=a22L_{3}=\dfrac{a^{2}}{2}: True. Correct answers: (1), (2), (3) and (4)
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