JEE Main 2016MathematicsLimitsMediumMCQ

JEE Main 2016 — Limits Question with Solution

JEE Main 2016 (03 Apr)

Question

Let P=limx→0+1+tan2x 12x, then logP is equal to

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Show full solutionCorrect option: A
Correct answer
A12

Step-by-step explanation

Let P=1+tan2x12x      (1∞ form)

If l=limx→afxgx, where limx→afx→1 and limx→agx→∞, then

l=elimx→afx-1·gx

⇒ P=elimx→0+1+tan2x-1·12x

⇒ P=elimx→0+12tanxx2

Using limx→0tanxx=0, we get

P=e12

⇒ logP=loge12=12.
 

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About this question

This is a previous-year question from JEE Main 2016, covering the Limits chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.