Jee mains 2016 Mathematics 52

Calculus Level 3

Let p = lim x 0 + ( 1 + tan 2 x ) 1 2 x \displaystyle p=\lim _{ x\rightarrow { 0 }^{ + } }{ { \left( 1+\tan ^{ 2 }{ \sqrt { x } } \right) }^{ \frac { 1 }{ 2x } } } , compute ln ( p ) \ln { \left( p \right) } .

0.5 0.25 1 2

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1 solution

Tom Engelsman
May 4, 2021

Let us take:

ln ( p ) = lim x 0 + ln ( 1 + tan 2 x ) 2 x \large \ln(p) = \lim_{x \rightarrow 0^{+}}\frac{\ln(1+\tan^{2} \sqrt{x})}{2x} (i).

Applying L'Hopital's Rule yields:

tan x 2 x \large \frac{\tan \sqrt{x}}{2\sqrt{x}} (ii)

followed by second application gives:

ln ( p ) = lim x 0 + sec 2 x 2 = 1 2 . \large \ln(p) = \lim_{x \rightarrow 0^{+}} \frac{\sec^{2} \sqrt{x}}{2} = \boxed{\frac{1}{2}}.

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