Exists f
.
df(x)/dx = f(x)
lim{y→0} (f(x+y) − f(x)) / y = f(x)
lim{y→0} f(x+y) − f(x) − y f(x) = 0
lim{y→0} f(x+y) − (1 + y) f(x) = 0
lim{y→0} f(x+y) = lim{y→0} (1 + y) f(x)
(Algebraic limit theorem.)
Plus y
in the parameter, times y
on the right hand side. Assumption: f
involves exponentiation. Let e
be a positive real number.
lim{y→0} e^(x+y) = lim{y→0} (1 + y) e^x
Since this equation holds for all x
, pick x = −y
.
lim{y→0} e^(y−y) = lim{y→0} (1 + y) e^−y
lim{y→0} 1 = lim{y→0} (1 + y) / e^y
Exponentiate both sides by 1/y
.
lim{y→0} 1^(1/y) = lim{y→0} (1 + y)^(1/y) / e
1 = lim{y→0} (1 + y)^(1/y) / e
e = lim{y→0} (1 + y)^(1/y)
1 = lim{y→0} (1 + y) / a^y
holds for any real number a>0, so you can not derive constant a=e from that equation. What do you mean by assuming f involves exponentiation? If you define exponential function bydf(x)/dx=f(x), f(0)=1
, you have to derive all calculation rules for exponentiation using that definition. Or you have to define exponentiation by other means and prove that exponentiation satisfies the conditions for f. See my fork.