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Limits — JEE Maths practice question

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

Question
lim⁡x→0tan⁡([−π2]x2)−x2tan⁡([−π2])sin⁡2x\displaystyle\lim_{x\to 0}\dfrac{\tan\left([-\pi^{2}]x^{2}\right)-x^{2}\tan\left([-\pi^{2}]\right)}{\sin^{2}x}, where [⋅][\cdot] is GIF
Atan⁡10+10\tan 10+10
Btan⁡10−10\tan 10-10correct
C10−tan⁡1010-\tan 10
DNone
Solution
Step 1: π≈3.14\pi\approx 3.14, so π2≈9.87\pi^{2}\approx 9.87. Therefore −π2≈−9.87-\pi^{2}\approx -9.87 and [−π2]=−10[-\pi^{2}]=-10. Step 2: Substitute:
lim⁡x→0tan⁡(−10x2)−x2tan⁡(−10)sin⁡2x\lim_{x\to 0}\dfrac{\tan(-10x^{2})-x^{2}\tan(-10)}{\sin^{2}x}
 Step 3: Use tan⁡(−y)=−tan⁡y\tan(-y)=-\tan y:
=lim⁡x→0x2tan⁡10−tan⁡(10x2)sin⁡2x=\lim_{x\to 0}\dfrac{x^{2}\tan 10-\tan(10x^{2})}{\sin^{2}x}
 Step 4: Divide top and bottom by x2x^{2}:
=lim⁡x→0tan⁡10−tan⁡(10x2)x2sin⁡2xx2=\lim_{x\to 0}\dfrac{\tan 10-\dfrac{\tan(10x^{2})}{x^{2}}}{\dfrac{\sin^{2}x}{x^{2}}}
 Step 5: Use standard limits: tan⁡(10x2)x2→10\dfrac{\tan(10x^{2})}{x^{2}}\to 10 and sin⁡2xx2→1\dfrac{\sin^{2}x}{x^{2}}\to 1. Step 6: Combine:
=tan⁡10−101=tan⁡10−10=\dfrac{\tan 10-10}{1}=\tan 10-10
 Correct answer: (2) tan⁡10−10\tan 10-10
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