Does log(decimal!) Exist?

Calculus Level 5

Evaluate: lim x 1 log e ( ( 10.5 x ) ! ) log e ( ( 10.5 ) ! ) x 1 \lim _{ x\rightarrow 1 }{ \frac { \log _{ e }{ \left( \left( 10.5x \right) ! \right) -\log _{ e }{ \left( \left( 10.5 \right) ! \right) } } }{ x-1 } }

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The answer is 25.181.

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1 solution

Aditya Kumar
Sep 30, 2015

A = lim x 1 log e ( ( 10.5 x ) ! ) log e ( ( 10.5 ) ! ) x 1 = lim x 1 log e ( Γ ( 10.5 x + 1 ) ) log e ( ( 10.5 ) ! ) x 1 U s i n g L h o s p i t a l s r u l e , A = 10.5 ψ ( 11.5 ) = 10.5 { γ 2 l n 2 + k = 1 11 2 2 k 1 } = 25.181 ψ i s u s e d t o r e p r e s e n t d i g a m m a f u n c t i o n . γ i s E u l e r M a s h c h e r o n i c o n s t a n t . A=\lim _{ x\rightarrow 1 }{ \frac { \log _{ e }{ \left( \left( 10.5x \right) ! \right) -\log _{ e }{ \left( \left( 10.5 \right) ! \right) } } }{ x-1 } } \\ =\lim _{ x\rightarrow 1 }{ \frac { \log _{ e }{ \left( \Gamma \left( 10.5x+1 \right) \right) -\log _{ e }{ \left( \left( 10.5 \right) ! \right) } } }{ x-1 } } \\ Using\quad L'hospital's\quad rule,\\ A=10.5\psi (11.5)\\ \quad =10.5\left\{ -\gamma -2ln2+\sum _{ k=1 }^{ 11 }{ \frac { 2 }{ 2k-1 } } \right\} \\ \quad =25.181\\ \psi \quad is\quad used\quad to\quad represent\quad digamma\quad function.\\ \gamma \quad is\quad Euler-Mashcheroni\quad constant.

@Abhishek Bakshi see this and comment.

Aditya Kumar - 5 years, 8 months ago

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very easy question... just the properties of the digamma function..

Abhishek Bakshi - 5 years, 8 months ago

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