Observations on power, exponential and logarithmic functions

Power functions namely functions of the form $x^\alpha$ where $\alpha$ is a real number, the exponential function $e^x$ and the logarithmic function $\ln(x)$ appear all the time in mathematics. They may appear by themselves or as part of compound expressions such as $\frac{x^2}{x^3+1}$ , $\frac{1}{e^x+1}$ or $x\ln(x)$ . It is essential for any student of mathematics to have firm grasp of these functions and visualize the pattern in this collection of functions. In this article, we restrict our attention to $x \ge 0$ because $x^\alpha\notin \mathbb{R}$ if $x < 0$ for certain values of $\alpha$ . Also, $\ln(x)\notin\mathbb{R}$ if $x < 0$ .

For $\alpha > 0$ , $x^\alpha$ is a strictly increasing function. That is, for any $x_1, x_2 \ge 0$ with $x_1 < x_2$ , we have $x_1^\alpha < x_2^\alpha$ .
For $\alpha < 0$ , $x^\alpha$ is a strictly decreasing function. That is, for any $x_1, x_2 \ge 0$ with $x_1 < x_2$ , we have $x_1^\alpha > x_2^\alpha$ .
Of course, for $\alpha = 0$ , $x^\alpha$ is identically equal to $1$ , a constant.
$x^\alpha = 1$ for $x = 1$ for all values of $\alpha$ .
If $x\in(0,1)$ , the family of functions $\{x^\alpha\}_{\alpha\in\mathbb{R}}$ is decreasing. That is, if $\alpha_2 > \alpha_1$ , we have $x^{\alpha_2} < x^{\alpha_1}$ for each $x\in(0,1)$ .

If $x > 1$ , the family of functions $\{x^\alpha\}_{\alpha\in\mathbb{R}}$ is increasing. That is, if $\alpha_2 > \alpha_1$ , we have $x^{\alpha_2} > x^{\alpha_1}$ for each $x > 1$ .
As , the function grows faster than any for any no matter how big is. That is, as for no matter how big is.
- In fact, $\beta^x - x^\alpha \rightarrow\infty$ as $x\rightarrow\infty$ for $\alpha > 0$ and $\beta > 1$ no matter how big $\alpha$ is and how small $\beta$ is. For example, $\lim\limits_{x\to \infty}\frac{x^{100000000000000000000000000}}{1.00000001^x} = 0$ .
As $x\rightarrow\infty$ , the function $\ln(x)\rightarrow\infty$ . However, as $x\rightarrow\infty$ , $\ln(x)$ (or, more generally,
$\log^x_{\beta}$ ) grows slower than any $x^\alpha$ for any $\alpha > 0$ (and for any $\beta > 1$ ) no matter how small $\alpha$ is (and how small $\beta$ is). That is, $x^\alpha > \log^x_{\beta}$ and in fact, $x^\alpha-\log^x_{\beta}\rightarrow\infty$ as $x\rightarrow\infty$ for $\alpha > 0$ and $\beta > 1$ no matter how small $\alpha$ is and how small $\beta$ is.

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