Unizor - Limits - F(x)=e^x

In the chapter describing exponential functions we introduced the concept of steepness β of an exponential function F(x)=a^x at point x=0 as a limit of a ratio between increment of a function between points x=0 and x=1/n (that is F(1/n)−F(0)=a^(1/n)−a^0) and a corresponding increment of an argument (that is 1/n−0) when n→∞, so a point x=1/n is getting closer and closer to a point x=0.
After a trivial simplification (since a^0=1) the steepness of a function F(x)=a^x at point x=0 is defined as
β = lim[n→∞](a^(1/n)−1)/(1/n) = lim[n→∞]n·(a^(1/n)−1)

At this point we would like you to recall the material presented in the lecture Algebra - Exponential Functions - Problems 2 about a steepness of exponential functions 2^x and 3^x at a point x=0. Refer to that lecture for refreshing.

We have proved there that the steepness of a function 2^x at point x=0 is less than 1, while the steepness of a function 3^x at point x=0 is greater than 1.
Assuming continuity of steepness as it depends on the base of the exponential function, it is reasonable to expect that somewhere between 2 and 3 there is a value of a base of an exponential function with the steepness equaled exactly to 1. This value of a base is denoted as e in mathematics and its approximate value is 2.71.
Therefore, we can write:
lim[n→∞] (e^(1/n)−1)/(1/n) = 1

Slightly deviating from the rigorousness we used to apply to our statements, we can derive from the last equality that there is an approximation
(e^(1/n)−1)/(1/n) ≅ 1
which becomes better and better as n→∞.
Or, resolving this for e,
e^(1/n)−1 ≅ 1/n or
e^(1/n) ≅ 1+1/n or
e ≅ (1+1/n)^n
Returning to exact equality, we can write the value of e as the following limit:
e = lim[n→∞] (1+1/n)^n

Summarizing the above logic, we have determined a specific value of the base of the exponential function with the steepness of 1 at point x=0. This value is designated a symbol e and is equal to
e = lim[n→∞] (1+1/n)^n
Thus, we have defined an exponential function
F(x) = e^x
Obviously, all properties of exponential functions are preserved in this particular function, including but not limited to:
e^0 = 1
e^1 = e
e^(x+y) = e^x·e^y
e^(x·y) = (e^x)^y

Another interesting fact is that steepness, as we introduced it for function F(x)=e^x at point x=0, can be defined at any other point using exactly the same methodology.
Consider any point x=x0. Now step forward from this point by a value 1/n to a point x=x0+1/n. Construct a ratio between the difference in the values of a function in these two points and the difference in the values of an argument:
β = [F(x0+1/n) − F(x0)] / (1/n)
As n→∞, the limit of this ratio characterizes the steepness of a function F(x) at point x0.

In case of the function we have introduced in this lecture, F(x)=e^x, the steepness at any point x0 equals to a limit of the following expression, as n→∞:
[e^(x0+1/n) − e^x0] / (1/n) =
= [e^(x0)·e^(1/n) − e^(x0)] / (1/n) =
= e^(x0)·[e^(1/n) − 1] / (1/n)
As n→∞, e^(x0) remains a constant while the expression
[e^(1/n) − 1] / (1/n), which represents the steepness of function ex at point x=0, has a limit of 1.
Therefore, the steepness of function e^x at any point x0 equals to a value of this function at this same point, that is e^(x0). This is quite a remarkable and unique property of function F(x)=e^x.