nice-calculus-prove-that-lim-x-xe-1-x-e-1-x-1-x-2-euler-mascheroni-constant-

Question Number 124831 by mnjuly1970 last updated on 06/Dec/20

\dots n i c e c a l c u l u s \dots

p r o v e t h a t ::

{l i m}_{x \to \infty} {{x e}^{\frac{1}{x}} - e^{\frac{- 1}{x}} Γ (\frac{1}{x})} = 2 + γ

γ : e u l e r - m a s c h e r o n i c o n s t a n t

Answered by Dwaipayan Shikari last updated on 06/Dec/20

\lim_{x \to \infty} ({x e}^{\frac{1}{x}} - e^{- \frac{1}{x}} x (\frac{e^{- \frac{γ}{x}} \prod^{\infty} e^{\frac{1}{n x}}}{\prod^{\infty} (1 + \frac{1}{n x})})) e^{\frac{1}{n x}} = 1 + \frac{1}{n x}

= \lim_{x \to \infty} ({x e}^{\frac{1}{x}} - {x e}^{- \frac{1}{x} - \frac{γ}{x}}) = x (1 + \frac{1}{x}) - x (1 - \frac{1}{x} - \frac{γ}{x})

= 1 + 1 + γ = 2 + γ \lim_{z \to 0} e^{z} = 1 + z

Commented by mnjuly1970 last updated on 06/Dec/20

t h a n k y o u

e x c e l l e n t \dots .

Commented by Dwaipayan Shikari last updated on 06/Dec/20

G r e a t s i r! :)

Answered by mindispower last updated on 06/Dec/20

Γ (\frac{1}{x}) = x Γ (1 + \frac{1}{x})

e^{\frac{1}{x}} = 1 + \frac{1}{x} + o (\frac{1}{x}), e^{- \frac{1}{x}} = (1 - \frac{1}{x} + o (\frac{1}{x}))

Γ (1 + \frac{1}{x}) = 1 + \frac{Ψ (1)}{x} + o (\frac{1}{x})

\Leftrightarrow x (1 + \frac{1}{x} + o (\frac{1}{x})) - (1 - \frac{1}{x} + o (\frac{1}{x})) . x (1 - \frac{γ}{x} + o (\frac{1}{x}))

= x + 1 + o (1) - (x - γ - 1 + o (1))

= γ + 2 + o (1)

\lim_{x \to 0} (γ + 2 + o (1)) = γ + 2

Commented by mnjuly1970 last updated on 06/Dec/20

t h a n k s a l o t s i r P o w r . .

Answered by mnjuly1970 last updated on 06/Dec/20

\frac{1}{x} \underset{t \to 0}{\overset{x \to \infty}{=}} t

{l i m}_{t \to 0} [\frac{1}{t} e^{t} - e^{- t} Γ (t)]

= {l i m}_{t \to 0} (\frac{e^{t} - e^{- t} Γ (t + 1)}{t})

\underset{r u l e}{\overset{h o p i t a l}{=}} {l i m}_{t \to 0} (\frac{e^{t} + e^{- t} Γ (t + 1) - Γ^{'} (t + 1) e^{- t}}{1})

= 1 + 1 - ψ (1) = 2 - (- γ) = 2 + γ ✓