More Powers Above

Calculus Level 5

For positive integers A , B , C , D A,B,C,D , we have

lim ⁡ x → 2 x x x x − 2 16 x x x − 16 \large \displaystyle \lim_{x \to 2} \frac {x^{x^{x^x}} - 2^{16}}{x^{x^x} - 16}

equals to

2 A 1 + B ( ln ⁡ 2 ) 2 + C ⋅ ( ln ⁡ 2 ) + 2 D ⋅ ln ⁡ 2 \frac {2^A}{1 + B ( \ln 2)^2 + C \cdot ( \ln 2) } + 2^D \cdot \ln 2

What is the value of A + B + C − D A + B + C - D ?


The answer is 2.

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Pi Han Goh by Pi Han Goh , 30, Malaysia

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1 solution

Ronak Agarwal
Oct 4, 2014

Applying L-Hopital's rule we get :

lim ⁡ x → 2 x x x x . x x x . l n ( x ) . ( 1 x . l n ( x ) + x ( x − 1 ) + x x l n ( x ) . ( l n ( x ) + 1 ) ) x x x ( x x − 1 + x x . l n ( x ) . ( 1 + l n ( x ) ) ) \LARGE \displaystyle \lim _{ x\rightarrow 2 }{ \frac { { x }^{ { x }^{ { x }^{ x } } }.{ x }^{ { x }^{ x } }.ln(x).(\frac { 1 }{ x.ln(x) } +{ x }^{ (x-1) }+{ x }^{ x }ln(x).(ln(x)+1)) }{ { x }^{ { x }^{ x } }({ x }^{ x-1 }+{ x }^{ x }.ln(x).(1+ln(x))) } }

Put the x = 2 x=2 to get :

2 16 . l n ( 2 ) . ( 1 2. l n ( 2 ) + 2 + 4. l n ( 2 ) . ( 1 + l n ( 2 ) ) ) 2 + 4. l n ( 2 ) . ( 1 + l n ( 2 ) ) \Large \displaystyle { \frac { { 2 }^{ 16 }.ln(2).(\frac { 1 }{ 2.ln(2) } +2+4.ln(2).(1+ln(2))) }{ 2+4.ln(2).(1+ln(2)) } }

Which on simplifying we get :

2 16 . l n ( 2 ) + 2 14 1 + 2. l n ( 2 ) 2 + 2. l n ( 2 ) \Large \displaystyle { 2 }^{ 16 }.ln(2)+\frac { { 2 }^{ 14 } }{ 1+2.{ ln(2) }^{ 2 }+2.ln(2) }

Hence A = 14 , B = 2 , C = 2 , D = 16 \boxed{A=14,B=2,C=2,D=16}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Did it the same way. Is there a more elegant solution ?

Arif Ahmed - 6 years, 8 months ago

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