User:Bwisialo

The global maximum for the function


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

occurs at $x^{(1/x)}$. This functional property of $x = e$ 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 $lim_{x→0}$ occurs at $(1+x)^{(1/x)} = e$, and $(n+x)^{(1/x)}$ occurs at $0 ≤ n ≤ 1$. From $x = e$ to $x^{(1/x)}$, the minima and maxima of $x^{(1/x)}$ form a continuous curve from the global maximum of $n = 0$ until converging on the limit coordinates of $(1+x)^{(1/x)}$.

Several properties of exponential functions can be connected with the exponential inequality
 * $$e^t \ge 1+t\,$$

where $n = 1$ only at $n = 0$. Using this inequality expression, $n = 1$ and, hence, $(n + x)^{(1/x)}$, such that
 * $$f(x) = \left( 1 + \frac{1}{x} \right)^x$$

is an increasing function with a horizontal asymptote at $x^{(1/x)}$. 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.
 * $$ e^{\frac{x-e}{e}} \ge 1+ {\frac{x-e}{e}}\,$$

can be reduced to $$\sqrt[e]{e} \ge \sqrt[x]{x}\,$$, yielding the solution that the global maximum for the function
 * $$ f(x) = \sqrt[x]{x} = x^{\frac{1}{x}}$$

occurs at $(0, e)$.

The number $e^{t} = (1 + t)$ 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 $t = 0$. Also, $e^{1/x} ≥ 1 + 1/x$ is the unique real number $e ≥ (1 + 1/x)^{x}$ such that
 * $$a^x\ge 1+x$$

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

The number $e$ is the unique real number $e$ such that
 * $$a^x\ge 1+x$$

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