Range of e^x/([x]+1) ,x >0 and where [x] denotes the greatest integer ?

Jul 13, 2018

$f : \left(0 , + \infty\right) \to \left(\frac{1}{2} , + \infty\right)$

Explanation:

I assume $\left[x\right]$ is the smallest integer bigger than $x$. In the following answer, we'll use the notation $\left\lceil x \right\rceil$, called the ceiling function.

Let $f \left(x\right) = {e}^{x} / \left(\left\lceil x \right\rceil + 1\right)$. Since $x$ is strictly bigger than $0$, this means that the domain of $f$ is $\left(0 , + \infty\right)$.

As $x > 0$, $\left\lceil x \right\rceil > 1$ and since ${e}^{x}$ is always positive, $f$ is always strictly bigger than $0$ in its domain. It is important to note that $f$ is not injective and is also not continuous at the natural numbers. To prove this, let $n$ be a natural number:

${R}_{n} = {\lim}_{x \to {n}^{+}} f \left(x\right) = {\lim}_{x \to {n}^{+}} {e}^{x} / \left(\left\lceil x \right\rceil + 1\right)$

Because $x > n$, $\left\lceil x \right\rceil = n + 1$.

${R}_{n} = {e}^{n} / \left(n + 2\right)$

${L}_{n} = {\lim}_{x \to {n}^{-}} f \left(x\right) = {\lim}_{x \to {n}^{-}} {e}^{x} / \left(\left\lceil x \right\rceil + 1\right)$

Similarly, $\left\lceil x \right\rceil = n$.

${L}_{n} = {e}^{n} / \left(n + 1\right)$

Since the left and right sided limits are not equal, $f$ is not continuous at the integers. Also, $L > R$ for all $n \in \mathbb{N}$.

As $f$ is increasing in intervals bounded by the positive integers, the "smallest values" per interval will be as $x$ approaches the lower bound from the right.

Hence, the minimum value of $f$ is going to be

${R}_{0} = {\lim}_{x \to {0}^{+}} f \left(x\right) = {\lim}_{x \to {0}^{+}} {e}^{x} / \left(\left\lceil x \right\rceil + 1\right) = {e}^{0} / \left(0 + 2\right) = \frac{1}{2}$

This is the lower bound of the range of $f$.

While it isn't truly correct to say that $f$ is increasing, it is in the sense, asymptotically, it approaches infinity - as proved below:

${\lim}_{x \to \infty} f \left(x\right) = {\lim}_{x \to \infty} {e}^{x} / \left(\left\lceil x \right\rceil + 1\right)$

As $\left\lceil x \right\rceil \ge x$, there exists a $\delta < 1$ such that $\left\lceil x \right\rceil = x + \delta$:

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

Let $u = x + \delta + 1 \implies x = u - \delta - 1$.

$= {\lim}_{u \to \infty} {e}^{u - \delta - 1} / u = \left[{\lim}_{u \to \infty} {e}^{u} / u\right] \cdot \frac{1}{e} ^ \left(\delta + 1\right)$

${e}^{u}$ increases exponentially while $u$ does so linearly, meaning that

${\lim}_{u \to \infty} {e}^{u} / u = \infty$

$\therefore \left[{\lim}_{u \to \infty} {e}^{u} / u\right] \cdot \frac{1}{e} ^ \left(\delta + 1\right) = \infty \cdot \frac{1}{e} ^ \left(\delta + 1\right) = \infty$

$\therefore {\lim}_{x \to \infty} f \left(x\right) = \infty$

Therefore the range of $f$ is

$\text{Range} = \left(\frac{1}{2} , \infty\right)$

The interval is open on the left because $1 / 2$ is still $f \left(0\right)$, and as $x$ approaches ${0}^{+}$, $f \left(x\right)$ only approaches $1 / 2$; it never truly is equal.