User:Bwisialo

The global maximum of ${\sqrt[{x}]{x}}$ or x^(1/x) occurs at x = e. This functional property is continuous with the limit property, lim_x→0 (1+x)^(1/x) = e, as illustrated by the minima and maxima of (n+x)^(1/x) for 0 ≤ n ≤ 1.

The global maximum for the function

$f(x)={\sqrt[{x}]{x}}=x^{\frac {1}{x}}$

occurs at x = e. This functional property of x^(1/x) is continuous with the limit property of the function

$\lim _{x\to 0}\left(1+x\right)^{\frac {1}{x}}=e.$

The two functions are instances of

$f(x)=(n+x)^{\frac {1}{x}},$

where x^(1/x) occurs at n = 0, and (1+x)^(1/x) occurs at n = 1. From n = 0 to n = 1, the minima and maxima of (n + x)^(1/x) form a continuous curve from the global maximum of x^(1/x) until converging on the limit coordinates of (0, e).

Several properties of exponential functions can be connected with the exponential inequality

$e^{t}\geq 1+t\,$

where e^t = (1 + t) only at t = 0. Using this inequality expression, e^1/x ≥ 1 + 1/x and, hence, e ≥ (1 + 1/x)^x, such that

$f(x)=\left(1+{\frac {1}{x}}\right)^{x}$

The global maximum of ${\sqrt[{x}]{x}}$ occurs at x = e.

is an increasing function with a horizontal asymptote at y = e. More generally,

$\lim _{x\to \infty }\left(1+{\frac {n}{x}}\right)^{x}=e^{n}.$

Additionally, the above exponential inequality can be used solve Steiner's Problem.

$e^{\frac {x-e}{e}}\geq 1+{\frac {x-e}{e}}\,$

can be reduced to ${\sqrt[{e}]{e}}\geq {\sqrt[{x}]{x}}\,$, yielding the solution that the global maximum for the function

$f(x)={\sqrt[{x}]{x}}=x^{\frac {1}{x}}$

occurs at x = e.

e is the unique positive number a such that a^x ≥ x + 1 holds for all x and such that a^x = x + 1 if and only if x = 0.

The number e is the unique real number such that

$\left(1+{\frac {1}{x}}\right)^{x}<e<\left(1+{\frac {1}{x}}\right)^{x+1}$

for all positive x. Since it can be proved that

$\lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x}=\lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x+1}$

the inequality above provides a demonstration that this limit is e. Also, e is the unique real number a such that

$a^{x}\geq 1+x$

is true for all real x; and e is the unique positive number a such that a^x = x + 1 if and only if x = 0, with the result that e is the unique real number a where the slope of a^x at x = 0 is equal to 1.

The number e is the unique real number a such that

$a^{x}\geq 1+x$

is true for all real x; and e is the unique positive number a such that a^x = x + 1 if and only if x = 0. Also, the inequality e^x = 1 + x can be used to prove e is the unique real number such that

$\left(1+{\frac {1}{x}}\right)^{x}<e<\left(1+{\frac {1}{x}}\right)^{x+1}$

for all positive x. Since it can be also be proved that

$\lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x}=\lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x+1}$

the inequality expressions above provide a demonstration that this limit is e.