What the Question Is Really Asking
This problem is about limits and function behavior. A limit describes what a function does near a value, not necessarily exactly at that value.
"Very tiny positive $x$"
$x$ is greater than $0$ but extremely close to $0$.
$$x \to 0^+$$
"Very large $x$"
$x$ grows without bound.
$$x \to \infty$$
For each function the goal is to describe what the output does in both of these regimes.
1. Function-by-Function Behavior
$a(x) = 1^x$
Since $1$ raised to any power is always $1$, this function never changes.
As $x \to 0^+$
$$a(x) \to 1$$
As $x \to \infty$
$$a(x) \to 1$$
In words: the output is always $1$.
$b(x) = -x$
This function just takes the opposite of $x$.
As $x \to 0^+$
$-x$ is a tiny negative number close to $0$.
As $x \to \infty$
$-x$ becomes a very large negative number.
In words: near $0$ from the positive side, the output is slightly below $0$; for huge $x$, the output goes to negative infinity.
$d(x) = \dfrac{1}{x}$
For positive numbers very close to $0$, dividing by $x$ makes the output very large. For large $x$, $\frac{1}{x}$ becomes very small and approaches $0$. This is a standard example of one-sided and infinite limit behavior.
As $x \to 0^+$
$$d(x) \to \infty$$
As $x \to \infty$
$$d(x) \to 0$$
In words: tiny positive inputs make huge outputs; huge inputs make outputs very close to $0$.
$f(x) = \left(\dfrac{1}{x}\right)^x$
Rewrite as:
$$\left(\frac{1}{x}\right)^x = e^{x\ln(1/x)}$$
As $x \to 0^+$, the quantity $x\ln(1/x) \to 0$, so the whole expression approaches $e^0 = 1$. As $x \to \infty$, the base $\frac{1}{x}$ goes to $0$ and raising a tiny positive number to a very large power makes it go to $0$.
As $x \to 0^+$
$$f(x) \to 1$$
As $x \to \infty$
$$f(x) \to 0$$
In words: for very tiny positive $x$, the output is close to $1$; for very large $x$, the output is close to $0$.
$g(x) = \left(1 + \dfrac{1}{x}\right)^x$
This function is extremely important because it approaches the number $e$. One standard definition of $e$ is:
$$e = \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x$$
When $x \to 0^+$, the exponent $x$ becomes tiny while $x\ln(1+1/x) \to 0$, so the whole expression approaches $1$.
As $x \to 0^+$
$$g(x) \to 1$$
As $x \to \infty$
$$g(x) \to e \approx 2.71828$$
In words: near $0$ on the positive side, the output is close to $1$; for huge $x$, the output gets closer and closer to $e$.
$h(x) = e^x$
The function $e^x$ is called the natural exponential function. It plays a special role in calculus and exponential growth. For small positive $x$, $e^x$ is just a little bigger than $1$. For large $x$, it grows very quickly without bound.
As $x \to 0^+$
$$h(x) \to 1 \text{ from above}$$
As $x \to \infty$
$$h(x) \to \infty$$
In words: for very tiny positive $x$, the output is slightly greater than $1$; for very large $x$, the output becomes extremely large.
$k(x) = 1 + x$
This is a linear function.
As $x \to 0^+$
$$1 + x \to 1 \text{ from above}$$
As $x \to \infty$
$$1 + x \to \infty$$
In words: for very tiny positive $x$, the output is slightly greater than $1$; for very large $x$, the output becomes very large.
2. What Does $g(x)$ Have to Do with the Number $e$?
The function
$$g(x) = \left(1+\frac{1}{x}\right)^x$$
is directly connected to the constant $e$, because as $x$ becomes very large,
$$\left(1+\frac{1}{x}\right)^x \to e$$
This is one of the classic definitions of $e$, whose decimal value begins $2.718281828\ldots$
Simple explanation: $g(x)$ is a function whose outputs get closer and closer to $e$ as $x$ gets larger.
3. Relationship Between $h(x) = e^x$ and $k(x) = 1+x$ for Very Small Positive $x$
Near $x = 0$, the graph of $e^x$ has tangent line $y = 1 + x$. That is, $e^x$ has tangent line $x + 1$ at $(0, 1)$ with slope $1$. That is why, for very small values of $x$, $e^x$ and $1 + x$ are very close.
Another way to say this is that the beginning of the Maclaurin expansion of $e^x$ is:
$$e^x = 1 + x + \frac{x^2}{2!} + \cdots$$
so when $x$ is tiny, the higher-power terms are very small and $e^x \approx 1 + x$.
Conclusion
For very small positive $x$, $h(x) = e^x$ and $k(x) = 1 + x$ have almost the same value.
4. Quick Summary Table
Use this table as a quick reference for all seven functions:
| Function | As $x \to 0^+$ | As $x \to \infty$ |
|---|---|---|
| $1^x$ | $1$ | $1$ |
| $-x$ | tiny negative number close to $0$ | very large negative number |
| $\dfrac{1}{x}$ | very large positive number | $0$ |
| $\left(\dfrac{1}{x}\right)^x$ | $1$ | $0$ |
| $\left(1+\dfrac{1}{x}\right)^x$ | $1$ | $e$ |
| $e^x$ | slightly bigger than $1$ | very large positive number |
| $1+x$ | slightly bigger than $1$ | very large positive number |
The entries involving $e$, exponential functions, and limit language follow from standard definitions and properties of limits and exponential functions.
5. Final Written Answer
For this problem, "very tiny positive $x$" means $x \to 0^+$, so $x$ is greater than $0$ but very close to $0$, and "very large $x$" means $x \to \infty$.
For $a(x) = 1^x$, the output is always $1$. For $b(x) = -x$, the output is a tiny negative number near $0$ when $x$ is very small and positive, and it becomes very negative when $x$ is very large. For $d(x) = \frac{1}{x}$, the output becomes very large when $x \to 0^+$, and it approaches $0$ when $x \to \infty$.
For $f(x) = \left(\frac{1}{x}\right)^x$, the output approaches $1$ as $x \to 0^+$ and approaches $0$ as $x \to \infty$. For $g(x) = \left(1+\frac{1}{x}\right)^x$, the output approaches $1$ as $x \to 0^+$ and approaches $e$ as $x \to \infty$.
For $h(x) = e^x$, the output is slightly bigger than $1$ for very tiny positive $x$, and it grows without bound for very large $x$. For $k(x) = 1+x$, the output is also slightly bigger than $1$ for very tiny positive $x$, and it becomes very large for very large $x$.
Also, $g(x)$ is important because $\left(1+\frac{1}{x}\right)^x \to e$ as $x \to \infty$.
Finally, $h(x) = e^x$ and $k(x) = 1+x$ are very close for small positive $x$ because $1+x$ is the tangent line approximation to $e^x$ at $x = 0$.
6. Sources
- • OpenStax, Calculus Volume 1, section on exponential and logarithmic functions.
- • OpenStax, Calculus Volume 1, section on derivatives of exponential and logarithmic functions.
- • OpenStax, Calculus Volume 1, section on integrals, exponential functions, and logarithms.
- • Wolfram MathWorld, entry on e.
- • Wolfram MathWorld, entry on Maclaurin series.
- • Khan Academy, intro to limits and one-sided limits.