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Limits: Possible Value 3kx

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

Question
The possible value(s) of kk for which lim⁡x→∞2x3−(tan⁡−1x)38πx3cot⁡−1(∣kx∣)+k2x6sin⁡(1x3)−3kx3=12\displaystyle\lim_{x\to\infty}\dfrac{2x^{3}-(\tan^{-1}x)^{3}}{\dfrac{8}{\pi}x^{3}\cot^{-1}(|kx|)+k^{2}x^{6}\sin\left(\dfrac{1}{x^{3}}\right)-3kx^{3}}=\dfrac{1}{2} is
A00correct
B−1-1correct
C22correct
D44correct
Solution
Step 1: As x→∞x\to\infty: - tan⁡−1x→π2\tan^{-1}x\to\dfrac{\pi}{2}, so (tan⁡−1x)3→π38(\tan^{-1}x)^{3}\to\dfrac{\pi^{3}}{8}, a constant. So numerator ∼2x3\sim 2x^{3}. - cot⁡−1(∣kx∣)→0\cot^{-1}(|kx|)\to 0 if k≠0k\neq 0 (and if k=0k=0, cot⁡−10=π2\cot^{-1}0=\dfrac{\pi}{2}). - x6sin⁡1x3∼x6⋅1x3=x3x^{6}\sin\dfrac{1}{x^{3}}\sim x^{6}\cdot\dfrac{1}{x^{3}}=x^{3}. Step 2: Divide numerator and denominator by x3x^{3}. The numerator becomes 2−(tan⁡−1x)3x3→2−0=22-\dfrac{(\tan^{-1}x)^{3}}{x^{3}}\to 2-0=2. Step 3: For the denominator (after dividing by x3x^{3}):
8πcot⁡−1(∣kx∣)+k2x3sin⁡1x3−3k\dfrac{8}{\pi}\cot^{-1}(|kx|)+k^{2}x^{3}\sin\dfrac{1}{x^{3}}-3k
 Step 4: Use cot⁡−1(∣kx∣)=tan⁡−11∣kx∣∼1∣kx∣\cot^{-1}(|kx|)=\tan^{-1}\dfrac{1}{|kx|}\sim\dfrac{1}{|kx|} for k≠0k\neq 0, so 8πcot⁡−1(∣kx∣)→0\dfrac{8}{\pi}\cot^{-1}(|kx|)\to 0. Also k2x3sin⁡1x3→k2⋅1=k2k^{2}x^{3}\sin\dfrac{1}{x^{3}}\to k^{2}\cdot 1=k^{2}. Step 5: So denominator →0+k2−3k=k2−3k\to 0+k^{2}-3k=k^{2}-3k. Step 6: Set the limit equal to 12\dfrac{1}{2}:
2k2−3k=12  ⇒  k2−3k=4  ⇒  k2−3k−4=0\dfrac{2}{k^{2}-3k}=\dfrac{1}{2}\;\Rightarrow\;k^{2}-3k=4\;\Rightarrow\;k^{2}-3k-4=0
(k−4)(k+1)=0  ⇒  k=4 or k=−1(k-4)(k+1)=0\;\Rightarrow\;k=4\text{ or }k=-1
 Step 7: Special case k=0k=0. Then cot⁡−1(0)=π2\cot^{-1}(0)=\dfrac{\pi}{2}, so 8π⋅π2=4\dfrac{8}{\pi}\cdot\dfrac{\pi}{2}=4. Also k2x6sin⁡(⋅)=0k^{2}x^{6}\sin(\cdot)=0 and −3k⋅x3=0-3k\cdot x^{3}=0. After dividing by x3x^{3}: denominator →41⋅\to\dfrac{4}{1}\cdot... need to recheck. Actually for k=0k=0: numerator after /x3/x^{3} is 22, denominator after /x3/x^{3} is 8π⋅π2=4\dfrac{8}{\pi}\cdot\dfrac{\pi}{2}=4. So limit =24=12=\dfrac{2}{4}=\dfrac{1}{2}. Valid. So k=0k=0. Correct answers: (1) 00, (2) −1-1 and (4) 44
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