Maxima/Euler's Number

Euler's Number with Maxima

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• Euler's number e consideration.

f(dt):=(3^dt-1)/dt;

limit(f(dt), dt, 0);

(%o1)	f(dt):=(3^dt-1)/dt

(%o2)	log(3)

$\lim _{{dt}\to 0}{\cfrac {(3^{dt}-1)}{dt}}=log3$

kill (f,F,dt) $
ct:10 $
f(n,dt):=(n^dt-1)/dt;
for n:1 while n<=ct step 1 do (
    F[n] : [limit(f(n,dt), dt, 0)],
    print ("n=",n,"Result: ",float(F[n]))
) $

(%o81)	f(n,dt):=(n^dt-1)/dt
"n="" "1" ""Result: "" "[0.0]" "
"n="" "2" ""Result: "" "[0.6931471805599453]" "
"n="" "3" ""Result: "" "[1.09861228866811]" "
"n="" "4" ""Result: "" "[1.386294361119891]" "
"n="" "5" ""Result: "" "[1.6094379124341]" "
"n="" "6" ""Result: "" "[1.791759469228055]" "
"n="" "7" ""Result: "" "[1.945910149055313]" "
"n="" "8" ""Result: "" "[2.079441541679836]" "
"n="" "9" ""Result: "" "[2.19722457733622]" "
"n="" "10" ""Result: "" "[2.302585092994046

If n = e (2.718281828459045....) then the result is "1".

%e, numer;
f(n,dt):=(n^dt-1)/dt;
limit(f(%e,dt), dt, 0),numer;

(%o74)	2.718281828459045
(%o75)	f(n,dt):=(n^dt-1)/dt
(%o76)	1