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← JEE Main 2014 · 6 Apr · Shift 1
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JEE Main 2014 · 6 Apr · Shift 1, Mathematics question 70

lim⁡x→0sin⁡(πcos⁡2x)x2\displaystyle\lim_{x\to0}\frac{\sin\left(\pi\cos^2x\right)}{x^2} is equal to :

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Answer and worked solution

Official answer: B

sin⁡(πcos⁡2x)=sin⁡(π−πcos⁡2x)=sin⁡(πsin⁡2x)≈πsin⁡2x≈πx2\sin(\pi\cos^2x)=\sin(\pi-\pi\cos^2x)=\sin(\pi\sin^2x)\approx\pi\sin^2x\approx\pi x^2. So the limit is π\pi.

Topics:Limits
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