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?
math for fun
To see this working, head to your live site.
Jan 11, 2021
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
0
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