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Calculus Level 5

g n ( x ) = x x ⋯ n times ⋯ x h n ( x ) = ∑ r = 0 n x r r ! \Large{g_{n} (x) = {x^{x^{{\cdots \text{n times} \cdots}^x}}}}\\ \\ \Large{h_n (x) = \displaystyle \sum_{r=0}^{n} \dfrac{x^r}{r!}}

A = lim ⁡ x → ∞ lim ⁡ n → ∞ ( g n ( x ! ) ) ! × h n ( g n ( x ! ) ) g n ( x ! ) × ( g n ( x ! ) ) ( g n ( x ! ) ) \Large {A = \lim_{x\to\infty} \lim_{n\to\infty} \dfrac{\left(g_n(x!)\right)! × h_n\left(g_n(x!)\right)}{\sqrt{g_n(x!)} × {\left(g_n(x!)\right)}^{\left(g_n(x!)\right)}}}

Evaluate ⌊ A 2 ⌋ \lfloor A^2 \rfloor

Details and assumptions \text{Details and assumptions} :-

  • Here n ! n! is the factorial notation , n ! = n × ( n − 1 ) × ( n − 2 ) × ⋯ × 2 × 1 n! = n×(n-1)×(n-2)×\cdots×2×1 .
  • Here ⌊ . ⌋ \lfloor . \rfloor denotes the greatest integer function


The answer is 6.

Ashish Menon by Ashish Menon , 20, India

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1 solution

Ashish Menon
Feb 8, 2018

Let g n ( x ! ) = y g_n(x!) = y .
Now, lim ⁡ n → ∞ h n ( y ) = 1 + y + y 2 2 ! + y 3 3 ! ⋯ = e y \lim_{n\to\infty} h_n(y) = 1 \ + \ y \ + \ \dfrac{y^2}{2!} \ + \ \dfrac{y^3}{3!} \ \cdots = \ e^y .

A = lim ⁡ y → ∞ y ! × e y y × y y = lim ⁡ y → ∞ 2 π y × y e y × e y y × y y Using Stirling’s Approximation = 2 π ∴ ⌊ A 2 ⌋ = 6 \begin{aligned} A & = \lim{y\to \infty} \dfrac{y! × e^y}{\sqrt{y} × y^y}\\ \\ & = \lim{y\to \infty} \dfrac{\sqrt{2\pi y}×{\frac{y}{e}}^y × e^y}{\sqrt{y}×y^y} \ \text{Using Stirling's Approximation} \\ \\ & = \sqrt{2\pi}\\ \\ \therefore \lfloor A^2 \rfloor & = \color{#D61F06}{\boxed{6}} \end{aligned}

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Did it the same way!!!

Aaghaz Mahajan - 3 years, 4 months ago

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