
Deriving ln (x) series - Physics Forums
Dec 8, 2007 · How do you go about deriving the series expansion of ln (x)? 0 < x I got the representation at math.com but i'd still like to know how they got it. It's...
Derivative of 1/lnx - Physics Forums
Oct 28, 2007 · dy/dx = -1/xlnx Third Attempt: ln y = -lnx y = -x y' = -1 Which one is it? =/ Physics news on Phys.org First-of-its-kind measurement may help physicists learn about gluons, which …
Limit of x - lnx as x approaches infinity: Solving Indeterminate ...
May 11, 2010 · Which might be why you ended up with a value of 0 instead of the right value, which is infinity. It's much simpler to write x - lnx as x (1 - lnx/x), and note that 1 - lnx/x --> 1 as …
The limit as x approaches 1 of x / ln (x) - Physics Forums
Sep 6, 2008 · lim as x->1 (x/lnx) now me remove the natural lag lim as x->1 ( e^x/ x) so as X approaches 1 we get lim x->1 (e^1/1)=e :) What makes you think that you can remove the …
Understanding Why ln (x) and e Cancel Out - Physics Forums
Aug 4, 2012 · Hi, I have a hard time understanding why ln (x) and e cancel out, when, for example, we have something like: eln(2x+3) I tried an internet search but I did not get any …
L'Hopital on limit of tanx (lnx) as x ->0 (from the right).
Mar 20, 2012 · L'Hopital on limit of tanx(lnx) as x -->0 (from the right). Regarding this solution to the lim of (tanx)(lnx) as x approaches zero (from the right). I'm confused about the part I …
How to integrate (ln x) (squared) - Physics Forums
Feb 6, 2006 · Hi, I've been struggling with this problem for hours, so I was wondering if someone here could help me out, thanks: The problem is: How to integrate: (ln x)(to the power of 2) Thanks
How to prove ln (x) < sqrt (x) for all x>0 - Physics Forums
Sep 27, 2010 · Hi, I know I can use a graphical calculator to easily show that How to prove ln(x) < sqrt(x) for all x>0 But I wonder if there is a rigorous way to demonstrate this.
What is the integral 1/x log (x) or lnx? - Physics Forums
Oct 26, 2011 · Homework Statement I know it is supposed to be lnx however I find something peculiar. When I integrate it in wolfram alpha they give the integral as log(x). What the heck is …
What is the limit of x^lnx as x approaches 0+? - Physics Forums
Jan 22, 2012 · Well, I know that e^ln (x^lnx) is the same as x^lnx, but I put aside the e for the end of the equation. ln (x^lnx)= lnx*lnx. To apply L'Hospitals Rule I have to have it in quotient form, …