I am wondering what the limit of (ln(x!))/x would be as x approaches 0... I have tried to trace the graph with the help of Desmos and found it to be around -0.5772... I tried to solve and prove it myself but was unable to. When I finally used Wolfram Alpha, it told me that the answer is The Negative Euler's Mascheroni Constant. I'm wondering how to prove it. Can someone please help me out?
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Jan 11
Hey, can someone please help me evaluate this interesting limit??? (Limit as x approaches 0 of ((ln(x!))/x))
Hey, can someone please help me evaluate this interesting limit??? (Limit as x approaches 0 of ((ln(x!))/x))
2 answers2 replies
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Hello there..I think using the l'hospital rule might help.. Differentiating ln(x!) Leads us to the Euler mascheroni constant only...My opinion but ain't sure..
Hey there...! Thanks a lot for your reply! Yeah, differentiating both the sides definitely works but the problem is the functions rendered after applying the L'Hôpital's Rule are not elementary. I am actually looking for a creative way of solving it like how blackpenredpen himself solved the limit of ((n!)/(n^n))^(1/n) as n approaches infinity in this video ( https://www.youtube.com/watch?v=89d5f8WUf1Y ) I really loved it myself and would be really pleased if found a similar solution to this problem..... But your answer was pretty helpful too... Thank you!
No problems...Happy to help..
Yes I have seen that video...Its a really good method.. I am also trying to find a similar method which is creative and Diff from the complex l'hospital approach