Limited Trigonometry- I

Calculus Level 5

f ( x ) = t a n ( x 2 ) ⋅ s e c ( x ) + t a n ( x 2 2 ) ⋅ s e c ( x 2 ) + . . + t a n ( x 2 n ) ⋅ s e c ( x 2 n − 1 ) f(x)=tan\left( \frac{x}{2} \right) \cdot sec \left( x \right)+tan\left( \frac{x}{2^2} \right) \cdot sec \left( \frac{x}{2} \right)+..+tan\left( \frac{x}{2^n} \right) \cdot sec \left( \frac{x}{2^{n-1}} \right) g ( x ) = f ( x ) + t a n ( x 2 n ) g(x)=f(x)+tan\left( \frac{x}{2^n} \right)

where x ∈ ( − π 2 , π 2 ) x \in \left( -\frac{\pi}{2} , \frac{\pi}{2} \right) and n ∈ N n \in \mathbb{N}

Evaluate the limit : lim ⁡ x → 0 ( g ( x ) x ) 1 x \displaystyle \lim_{x \to 0} \left( \dfrac{g(x)}{x} \right)^{\frac{1}{x}}

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e 1 3 e^{\frac{1}{3}} 1 e Limit does not exist. None of the above.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Nishu Sharma
May 6, 2015

Well , According to Question framing , I orally conclude that It must be

f ( x ) = tan ⁡ x − tan ⁡ x 2 n g ( x ) = tan ⁡ x L = lim ⁡ x → 0 ( tan ⁡ x x ) 1 x = e lim ⁡ x → 0 ( tan ⁡ x − x x 2 ) = 1 f(x)=\tan { x } -\tan { \cfrac { x }{ { 2 }^{ n } } } \\ g(x)=\tan { x } \\ L=\lim _{ x\rightarrow 0 }{ { \left( \cfrac { \tan { x } }{ x } \right) }^{ \cfrac { 1 }{ x } } } ={ e }^{ \lim _{ x\rightarrow 0 }{ { \left( \cfrac { \tan { x } -x }{ { x }^{ 2 } } \right) } } }=1

Q.E.D

P r o o f Proof : Let concentrate on first term , we have to make it telescopic type !

T 1 = sin ⁡ x 2 cos ⁡ x 2 cos ⁡ x = sin ⁡ ( x − x 2 ) cos ⁡ x 2 cos ⁡ x = tan ⁡ x − tan ⁡ x 2 x → x 2 T 2 = tan ⁡ x 2 − tan ⁡ x 2 2 f ( x ) = ∑ 1 n T k = tan ⁡ x − tan ⁡ x 2 n \displaystyle{{ T }_{ 1 }=\cfrac { \sin { \cfrac { x }{ 2 } } }{ \cos { \cfrac { x }{ 2 } \cos { x } } } =\cfrac { \sin { (x-\cfrac { x }{ 2 } ) } }{ \cos { \cfrac { x }{ 2 } \cos { x } } } =\tan { x } -\tan { \cfrac { x }{ 2 } } \\ x\rightarrow \cfrac { x }{ 2 } \\ { T }_{ 2 }=\tan { \cfrac { x }{ 2 } } -\tan { \cfrac { x }{ { 2 }^{ 2 } } } \\ f(x)=\sum _{ 1 }^{ n }{ { T }_{ k } } =\tan { x } -\tan { \cfrac { x }{ { 2 }^{ n } } } }

Hence Prooved !

2 Helpful 0 Interesting 0 Brilliant 0 Confused

u could have used tan x expansion series!! makes things easy

A Former Brilliant Member - 6 years, 1 month ago

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Is my method is hard for you ? :P

well I did it in max 30-35 second's , So i don't think i used very big method .

So i don't think using expansion of tan x , can be solved in this time !

Nishu sharma - 6 years, 1 month ago

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it could be!! for sure! i did in tht way only!+ my method was just a secondary to urs!

A Former Brilliant Member - 6 years, 1 month ago

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