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.

# 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

4 Comments

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

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..