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← JEE Main 2016 · 3 Apr · Shift 1
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JEE Main 2016 · 3 Apr · Shift 1, Mathematics question 70

Let p=lim⁡x→0+(1+tan⁡2x)12xp=\displaystyle\lim_{x\to0+}\left(1+\tan^2\sqrt x\right)^{\frac1{2x}} then log p is equal to :

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Answer and worked solution

Official answer: C

log⁡p=lim⁡x→0+ln⁡(1+tan⁡2x)2x=lim⁡tan⁡2x2x=12.\log p=\displaystyle\lim_{x\to0^+}\frac{\ln(1+\tan^2\sqrt x)}{2x}=\lim\frac{\tan^2\sqrt x}{2x}=\frac12.

Topics:Limits
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