Why does the proof for d/dx lnx use Euler's constant?

[aclark609]

My question has to do with Euler's constant towards the end of the proof:

d/dx ln/x = lim h → 0 1/h ln(x+h) - lnx
= lim h→0 1/h ln[(x+h/x]
= lim h→0 1/h ln(1 + h/x)
= lim h→0 ln(1 + h/x)^(1/h)(1/x)(x/1)
= lim h→0 1/x ln(1+h/x)^(x/h) ;

let u = x/h

= 1/x lim h→0 ln(1+1/u)^u
= 1/x lne
= 1/x

Doesn't Euler's constant deal with the lim h→∞ instead of 0? Are they interchangeable? If so, why?

 

[Hypersphere]

My question has to do with Euler's constant towards the end of the proof:

d/dx ln/x = lim h → 0 1/h ln(x+h) - lnx
= lim h→0 1/h ln[(x+h/x]
= lim h→0 1/h ln(1 + h/x)
= lim h→0 ln(1 + h/x)^(1/h)(1/x)(x/1)
= lim h→0 1/x ln(1+h/x)^(x/h) ;

let u = x/h

= 1/x lim h→0 ln(1+1/u)^u
= 1/x lne
= 1/x

Doesn't Euler's constant deal with the lim h→∞ instead of 0? Are they interchangeable? If so, why?

If you let u→∞, you'll get e (which can be called Euler's number, Euler's constant usually means another number γ). Clearly, h→0 implies u→∞.

 

[aclark609]

EDIT: lim h-->0 should be lim u-->0

 

[Hypersphere]

EDIT: lim h-->0 should be lim u-->0

Why? When you make a substitution or a change of variables, you have to check how the new variable (i.e. u) depends on the original one (i.e. h). Otherwise, you'd get ridiculous results like the following (let y=2x)
2=\lim_{y\rightarrow 2} y = \lim_{x\rightarrow 2} 2x=4

EDIT: My point is that your final limit should be u\rightarrow \infty, not u\rightarrow 0.

 

[aclark609]

Aaaahhh. You're right. However if I allow h/x = u, then then lim u -->0 would be correct.