# Finest bounds for the Sinc hyperbolic function and applications

Abd Raouf Chouikha

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Abd Raouf Chouikha. Finest bounds for the Sinc hyperbolic function and applications. 2026. hal-05457921

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# Finest bounds for the Sinc hyperbolic function and applications

Abd Raouf Chouikha *

## Abstract

In this paper we prove for $x \in (0, \infty)$ the double inequality for the Sinc hyperbolic function

$$ \left( -2 \left( \frac{\tanh x}{x} \right)^{-\frac{r}{2}} + 3 \left( \cosh x \right)^{\frac{r}{2}} \right)^{\frac{1}{r}} < \frac{\sinh x}{x} < \cosh x \left( \frac{2}{3} \left( \frac{\tanh x}{x} \right)^{\frac{3}{2}r} + \frac{1}{3} \right)^{\frac{1}{r}}, $$

with equality when $r \to 0$.

Similar double inequalities are provided for the Huygens and The Wilker functions as well as for the Lazarewic function.

**Key Words and phrases:** Lazarewic inequalities; Huygens inequalities, Wilker inequalities, Sinc function ¹.

## 1 Introduction

The hyperbolic version of the Mitrinovic-Adamovic Inequality is given as for $x \neq 0$

$$ \cosh x < \left( \frac{\sinh x}{x} \right)^3 < \frac{\cosh x + 2}{3}. \quad (1) $$

The first one was obtained by Lazarewic and the second inequality is called hyperbolic Cusa-Huygens inequality.

There are many results on the refinement of the inequalities (1) in the literature, e.g, see

Recall Bhayo and Sandor [1] Theorem 1.5, proved the following for $x \geq 0$

$$ \exp \left( \frac{1}{2} \left( \frac{x}{\tanh x} - 1 \right) \right) < \frac{\sinh x}{x} < \exp \left( \left( \frac{x}{\tanh x} - 1 \right) \right). $$

As well as [1] Theorem 3.4

$$ \left( \frac{1 + \cosh x}{2} \right)^{2/3} < \frac{\sinh x}{x} < \left( \frac{1 + \cosh x}{2} \right). $$

<footnote>0*Universite Paris-Sorbonne, Paris-Nord, villetaneuse, 93430. juju153@yahoo.com.</footnote>

<footnote n="12010">12010 Mathematics Subject Classification: 26D07, 33B10, 26D05, 26D15</footnote>
Here is another bound provided by [6] for $x \ge 0$

$$ \left( \frac{1 + 2 \cosh x}{3} \right)^{1/2} < \frac{\sinh x}{x}. $$

On the other hand, [12] Theorem 2.2, proved for $x \ne 0$

$$ \cosh x < \left( \frac{\sinh x}{x} \right)^3 - \frac{12}{5} \left( 1 - \frac{x}{\sinh x} \right)^2. $$

This paper can be seen in a way as the hyperbolic version of [10]. Indeed, we will present new bounds for the Sinc hyperbolic function as well as for the Huygens and the Wilker hyperbolic functions in terms of one parameter functions $f(r, x)$ defined for $r \ge 0$ and for $x \in (0, \infty)$. These bounds appear to be finer than older

$$ \left( -2 \left( \frac{\tanh x}{x} \right)^{-\frac{3r}{2}} + 3 (\cosh x)^r \right)^{\frac{1}{3r}} < \left( \frac{\sinh x}{x} \right)^r < \\ \cosh x \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\sinh x}{x} \right)^{\frac{3r}{2}} (\cosh x)^{-\frac{3r}{2}} \right)^{\frac{1}{r}}. \quad (2) $$

More precisely, notice that when $r$ tends towards 0 one prove

$$ \lim_{r \to 0} \left( -2 \left( \frac{\tanh x}{x} \right)^{-\frac{3r}{2}} + 3 (\cosh x)^r \right)^{\frac{1}{3r}} = \\ \lim_{r \to 0} \cosh x \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\sinh x}{x} \right)^{\frac{9r}{2}} (\cosh x)^{-\frac{3r}{2}} \right)^{\frac{1}{r}} = \left( \frac{\sinh x}{x} \right). $$

Therefore if we bound $\frac{\sinh x}{x}$ by two functions $f(x)$ and $g(x)$:

$$ f(x) < \left( \frac{\sinh x}{x} \right) < g(x), $$

then there exists $r_0$ such that for $r < r_0$ one has

$$ f(x) < \left( -2 \left( \frac{\tanh x}{x} \right)^{-\frac{3r}{2}} + 3 (\cosh x)^r \right)^{\frac{1}{3r}}, \\ g(x) > \cosh x \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\sinh x}{x} \right)^{\frac{9r}{2}} (\cosh x)^{-\frac{3r}{2}} \right)^{\frac{1}{r}}. $$

Thus (2) appears to be the finest bound for the Sinc function.

We also prove analog bounds for Huygens and Wilker functions as well as for the Lazarewic function.
## 2 Main results

Next results give bounds for $\frac{\tanh x}{x}$ as well for the sinc function $\frac{\sinh x}{x}$

**Theorem 1.1** For $x \in (0, \infty)$ the function $\frac{\tanh x}{x}$ satisfies the following double inequality for any $r > 0$

$$ \left(-2 \left(\frac{\tanh x}{x}\right)^{-\frac{1}{2}r} + 3\right)^{\frac{1}{r}} < \left(-2 + 3 \left(\frac{\tanh x}{x}\right)^{\frac{1}{3}r}\right)^{\frac{1}{r}} < \\ \frac{\tanh x}{x} < \left(\frac{2}{3} \left(\frac{\tanh x}{x}\right)^{\frac{3}{2}r} + \frac{1}{3}\right)^{\frac{1}{r}}. $$

Moreover, the equalities are attained when $r$ tends towards 0.

**Theorem 1.2** For $x \in (0, \infty)$ the sinc function satisfies the following double inequality for any $r > 0$

$$ (i) \quad \left(-2 \left(\frac{\sinh x}{x}\right)^{-\frac{r}{2}} + 3\right)^{\frac{1}{r}} < \left(-2 + 3 \left(\frac{\sinh x}{x}\right)^{\frac{r}{3}}\right)^{\frac{1}{r}} < \frac{\sinh x}{x} < \left(\frac{2}{3} \left(\frac{\sinh x}{x}\right)^{\frac{3r}{2}} + \frac{1}{3}\right)^{\frac{1}{r}}. $$

$$ (ii) \quad \cosh x \left(-2 \left(\frac{\tanh x}{x}\right)^{-\frac{1}{2}r} + 3\right)^{\frac{1}{r}} < \frac{\sinh x}{x} < \left(\frac{(\cosh x)^r}{3} + \frac{2(\cosh x)^{\frac{1}{2}r}}{3} \left(\frac{\sinh x}{x}\right)^{\frac{3r}{2}}\right)^{\frac{1}{r}}. $$

Moreover, the equalities are attained when $r$ tends towards 0.

**Remark 1.2.1** It seems the following is true

$$ \cosh x \left(-2 \left(\frac{\tanh x}{x}\right)^{-\frac{1}{2}r} + 3\right)^{\frac{1}{r}} < \left(-2 \left(\frac{\tanh x}{x}\right)^{-\frac{r}{2}} + 3 (\cosh x)^{\frac{r}{3}}\right)^{\frac{1}{r}} < \frac{\sinh x}{x}. $$

If so we then can improve Theorem 1.2.

We may also derive bounds for the Huygens function

**Theorem 1.3** For $x \in (0, \infty)$ the following double inequality holds for any $r > 0$

$$ (2 \cosh x + 1) \left( 3 - 2 \left( \frac{\tanh x}{x} \right)^{-\frac{1}{2}r} \right)^{\frac{1}{r}} < \frac{2 \sinh x + \tanh x}{x} < $$
$$ (2 \cosh x + 1) \left( \frac{2}{3} \left( \frac{\tanh x}{x} \right)^{\frac{3}{2} r} + \frac{1}{3} \right)^{\frac{1}{r}} . $$

Moreover, the equalities are attained when $r$ tends towards 0.

By the same way one obtains bounds for the Wilker function

**Theorem 1.4** For $x \in (0, \infty)$ the following double inequality holds for any $r > 0$

$$ \begin{aligned} & (\cosh x)^2 \left( -2 \left( \frac{\tanh x}{x} \right)^{-\frac{1}{2}r} + 3 \right)^{\frac{2}{r}} + \left( -2 \left( \frac{\tanh x}{x} \right)^{-\frac{1}{2}r} + 3 \right)^{\frac{1}{r}} < \\ & \qquad \left( \frac{\sinh x}{x} \right)^2 + \frac{\tanh x}{x} < \\ & (\cosh x)^2 \left( \frac{2}{3} \left( \frac{\tanh x}{x} \right)^{\frac{3}{2}r} + \frac{1}{3} \right)^{\frac{2}{r}} + \left( \frac{2}{3} \left( \frac{\tanh x}{x} \right)^{\frac{3}{2}r} + \frac{1}{3} \right)^{\frac{1}{r}}. \end{aligned} $$

Moreover, the equalities are attained when $r$ tends towards 0.

Concerning the Lazarewic Inequality we prove the following

**Theorem 1.5** For $x \in (0, \infty)$ the following double inequality holds for any $r > 0$

$$ (i) \quad \left( -2 \left( \frac{1}{\cosh x} \left( \frac{\sinh x}{x} \right)^3 \right)^{-\frac{r}{2}} + 3 \right)^{\frac{1}{r}} < \left( -2 + 3 \left( \frac{1}{\cosh x} \left( \frac{\sinh x}{x} \right)^3 \right)^{\frac{r}{3}} \right)^{\frac{1}{r}} < \\ \qquad \frac{1}{\cosh x} \left( \frac{\sinh x}{x} \right)^3 < \left( \frac{2}{3} \left( \frac{1}{\cosh x} \left( \frac{\sinh x}{x} \right)^3 \right)^{\frac{3r}{2}} + \frac{1}{3} \right)^{\frac{1}{r}}. $$

For $x \in (0, \frac{\pi}{2})$ the following double inequality holds for any $r > 0$

$$ (ii) \quad \left( 3 - 2 \left( \frac{\sinh x}{x} \right)^{-\frac{3r}{2}} \left( \cosh x \right)^{\frac{r}{2}} \right)^{\frac{1}{r}} < \frac{1}{\cosh x} \left( \frac{\sinh x}{x} \right)^3 < \\ \qquad \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\sinh x}{x} \right)^{\frac{9r}{2}} \left( \cosh x \right)^{-\frac{3r}{2}} \right)^{\frac{1}{r}}. $$

Moreover, the equalities are attained when $r$ tends towards 0.
### 3 Proofs

The following lemmas will be useful in the sequel, [5], [11]

**Lemma 3.1** For $x \in (0, \infty)$, and $p, q \ge 0$ the following inequalities holds

$$ (i) \quad \left( \frac{\sinh x}{x} \right)^p > 1 + \frac{px^2}{6} + \left( \frac{1}{120} p + \frac{1}{72} p(p-1) \right) x^4 + \left( \frac{1}{5040} p + \frac{1}{720} p(p-1) + \frac{1}{1296} p(p-1)(p-2) \right) x^6, $$

$$ (ii) \quad 1 - \frac{qx^2}{3} + \left( \frac{2q}{15} + \frac{q(q-1)}{18} \right) x^4 + \left( -\frac{17q}{315} - \frac{2q(q-1)}{45} - \frac{q(q-1)(q-2)}{162} \right) x^6 < \left( \frac{\tanh x}{x} \right)^q, $$

$$ (iii) \quad \left( \frac{\log x}{\sinh x} \right) > -\frac{1}{6}x^2 + \frac{1}{180}x^4 - \frac{1}{2835}x^6 + \frac{1}{37800}x^8, $$

$$ (iv) \quad \log \left( \frac{\sinh x}{x} \right) < \frac{1}{6}x^2 - \frac{1}{180}x^4 + \frac{1}{2835}x^6, \quad \log \left( \frac{\tanh x}{x} \right) < -\frac{1}{3}x^2 + \frac{7}{90}x^4 - \frac{62}{2835}x^6. $$

**Lemma 3.2** (Proposition 1 of [14]) Let the series $f(x) = \sum_{k=1}^{\infty} (-1)^{k+1} A(k)x^{2k}$ converge for $x \in (0, c)$, $c \in IR^+$. Suppose that the following statements are true:

(i) If $c < 1$, then the sequence $A(k)$, $k \in IN$ is a positive decreasing sequence that converges to 0.

(ii) If $c \ge 1$, then the sequence $A(k)$, $k \in IN$ is a positive sequence, $\lim_{k \to \infty} c^{2k}A(k) = 0$ and $A(k) > c^2A(k+1)$ for $k \ge 1$. Then, for all $x \in (0, c)$ and for all $n \in N$ and $m \in N$, we have

$$ \sum_{k=1}^{2n} (-1)^{k+1} A(k) x^{2k} < f(x) < \sum_{k=1}^{2n+1} (-1)^{k+1} A(k) x^{2k}. $$

However, in [13] Theorem 2, the following result has been proved:

**Lemma 3.3** Suppose $f$ is a real function on $(a, b)$, $n$ is a positive integer such that $f^k(a+), f^k(b-), k = 0, 1, 2, \dots$, $n$ exist and $f^n(x)$ is increasing on $(a, b)$, then for all $x \in (a, b)$ the following inequality holds:

$$ \sum_{k=0}^{n-1} \frac{f^k(b-)}{k!} (x-a)^k + \frac{1}{(a-b)^n} \left( f(b-) - \sum_{k=0}^{n-1} \frac{(a-b)^k f^k(a+)}{k!} \right) (x-b)^n \\ > f(x) > \sum_{k=0}^{n} \frac{f^k(a+)}{k!} (x-a)^k. $$
Furthermore, if $f^n(x)$ is decreasing on $(a, b)$ then the reversed inequalities of
hold.

**Lemma 3.4** ([10] Lemma 3.3) Let a positive $C^1$-function $u(x)$ defined for $x \in (a,b)$ such that $u(0) = 1$ then for $r > 0$ the following inequalities hold

$$
(i) \quad \left(-2(u(x))^{-\frac{r}{2}} + 3\right)^{\frac{1}{r}} < \left(-2 + 3(u(x))^{\frac{r}{3}}\right)^{\frac{1}{r}} < u(x) < \left(\frac{2}{3}(u(x))^{\frac{3r}{2}} + \frac{1}{3}\right)^{\frac{1}{r}}.
$$

$$
(ii) \quad \left(-2 + 3(u(x))^{\frac{2r}{3}}\right)^{\frac{1}{r}} < (u(x))^2 < \left(\frac{2}{3}(u(x))^{\frac{3r}{2}} + \frac{1}{3}\right)^{\frac{2}{r}} < \left(\frac{2}{3}(u(x))^{3r} + \frac{1}{3}\right)^{\frac{1}{r}}.
$$

$$
(iii) \quad \left(-2 + 3(u(x))^r\right)^{\frac{1}{r}} < (u(x))^3 < \left(\frac{2}{3}(u(x))^{\frac{3r}{2}} + \frac{1}{3}\right)^{\frac{3}{r}} < \left(\frac{2}{3}(u(x))^{\frac{9r}{2}} + \frac{1}{3}\right)^{\frac{1}{r}}.
$$

More generally, for $k > 0$ one gets

$$
(iv) \quad \left(-2 + 3 \left(u(x)\right)^{\frac{kr}{3}}\right)^{\frac{1}{r}} < \left(u(x)\right)^{k} < \left(\frac{2}{3} \left(u(x)\right)^{\frac{3kr}{2}} + \frac{1}{3}\right)^{\frac{1}{r}}.
$$

Moreover, all these inequalities become equalities when $r$ tends towards 0.

**Proof of Theorem 1.1**

We will now separately prove the double inequality.

(i) - Let us prove the right one inequality

$$
\frac{\tanh x}{x} < \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\tanh x}{x} \right)^{\frac{3r}{2}} \right)^{\frac{1}{r}}.
$$

Since $\frac{\tanh x}{x} < 1$ we have now to prove

$$
\left(\frac{\tanh x}{x}\right)^r - \frac{1}{3} - \frac{2}{3}\left(\frac{\tanh x}{x}\right)^{\frac{3r}{2}} < 0.
$$

It is easy to see that the function

$$
Z - \frac{1}{3} - \frac{2}{3} Z^{3/2} = -\frac{1}{3} (2\sqrt{Z} + 1)(\sqrt{Z} - 1)^2 < 0,
$$

$$
\text{where } Z = \left( \frac{\tanh x}{x} \right)^r < 1.
$$

(ii) - The left inequality of Theorem 1.1

$$
\left(-2 \left(\frac{\tanh x}{x}\right)^{-\frac{1}{2}r} + 3\right)^{\frac{1}{r}} < \frac{\tanh x}{x}
$$
is equivalent to

$$-2 \left( \frac{\tanh x}{x} \right)^{-\frac{1}{2}r} + 3 < \left( \frac{\tanh x}{x} \right)^{r}.$$

Notice that if we take $Z = (\frac{\tanh x}{x})^{\frac{1}{2}}$ then

$$-2 \left( \frac{\tanh x}{x} \right)^{-\frac{1}{2}r} + 3 - \left( \frac{\tanh x}{x} \right)^{r} = -\frac{2}{Z} + 3 - Z^2 = -\frac{(Z+2)(Z-1)^2}{Z} < 0,$$

since $Z < 1$. This proves the left inequality.

(iii) - Turn now to the last part of Theorem 1.1

The logarithm yields :

$$\ln \left( -2 \left( \frac{\tanh x}{x} \right)^{-\frac{3r}{2}} + 3 (\cosh x)^r \right)^{\frac{1}{3r}} = \frac{\ln \left( -2 \left( \frac{\tanh x}{x} \right)^{-\frac{3r}{2}} + 3 (\cosh x)^r \right)}{3r} = \\ \frac{1}{3r} \left( 3 \ln \left( \frac{\sinh x}{x} \right) r \right) + O(r) = \ln \left( \frac{\sinh x}{x} \right) + O(r).$$

Thus

$$\lim_{r \to 0} \left( -2 \left( \frac{\tanh x}{x} \right)^{-\frac{3r}{2}} + 3 (\cosh x)^r \right)^{\frac{1}{3r}} = \frac{\sinh x}{x}.$$

By the same way, let us take the logarithm

$$\ln \cosh x \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\tanh x}{x} \right)^{\frac{3r}{2}} \right)^{\frac{1}{r}} = \frac{1}{r} \ln \cosh x \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\tanh x}{x} \right)^{\frac{3r}{2}} \right) = \ln \left( \frac{\sinh x}{x} \right) + O(r).$$

Thus

$$\lim_{r \to 0} \cosh x \left( \frac{1}{3} + \frac{2}{3} \left( \frac{\tanh x}{x} \right)^{\frac{3r}{2}} \right)^{\frac{1}{r}} = \frac{\sinh x}{x}.$$

## Proofs of Theorems 1.2, 1.3 and 1.4

These Theorems are in fact an easy consequence of Theorem 1.1. Indeed, we may write for Theorem 1.2 (ii)

$$\frac{\sinh x}{x} = \cosh x \frac{\tanh x}{x}.$$

(i) of Theorem 1.2 is derived from (i) Lemma 3.4 where $u(x) = \frac{\sinh x}{x}$. For Theorem 1.3 we have

$$\frac{2 \sinh x + \tanh x}{x} = \frac{2 \cosh x \tanh x}{x} + \frac{\tanh x}{x} = (2 \cosh x + 1) \frac{\tanh x}{x}.$$
For Theorem 1.4 we have

$$ \left( \frac{\tanh x}{x} \right)^2 + \frac{\tanh x}{x} = (\cosh x)^2 \left( \frac{\tanh x}{x} \right)^2 + \frac{\tanh x}{x}. $$

### Proof of Theorem 1.5

(i) is derived from Lemma 3.3 (i) where $u(x) = \frac{1}{\cosh x} \left(\frac{\sinh x}{x}\right)^3$.

To prove (ii) consider the series representations of hyperbolic functions [11] or [5]:

$$ \left( \frac{1}{\cosh x} \left( \frac{\sinh x}{x} \right)^3 \right)^r > 1 + \frac{1}{15} r x^4 - \frac{4}{189} r x^6 + \left( \frac{1}{150} r + \frac{1}{450} r^2 \right) x^8 + \\ \left( -\frac{68}{31185} r - \frac{4}{2835} r^2 \right) x^{10} + \left( \frac{42842}{58046625} r + \frac{5969}{8930250} r^2 + \frac{1}{20250} r^3 \right) x^{12}, \\ -2 \left( \frac{\sinh x}{x} \right)^{-\frac{3r}{2}} (\cosh x)^{1/2} r + 3 < \left( 1 + \frac{1}{15} r x^4 - \frac{4}{189} r x^6 + \left( \frac{1}{150} r - \frac{1}{900} r^2 \right) x^8 + \right. \\ \left. \left( -\frac{68}{31185} r + \frac{2}{2835} r^2 \right) x^{10} \right). $$

Therefore, after regrouping and simplifying and using *Maple* it yields for the left inequality of (i):

$$ \left( \frac{1}{\cosh x} \left( \frac{\sinh x}{x} \right)^3 \right)^r - \left( -2 \left( \frac{\sinh x}{x} \right)^{-\frac{3r}{2}} (\cosh x)^{\frac{r}{2}} + 3 \right) > \\ \left( \frac{42842}{58046625} r + \frac{5969}{8930250} r^2 + \frac{1}{20250} r^3 \right) x^{12} - \frac{2}{945} x^{10} r^2 + \frac{1}{300} x^8 r^2 > 0. $$

For the right inequality of (ii) let us estimate

$$ \left( \frac{(\sinh x)^3}{\cosh xx^3} \right)^r < 1 + \frac{1}{15} r x^4 - \frac{4}{189} r x^6 + \left( \frac{1}{150} r + \frac{1}{450} r^2 \right) x^8 + \left( -\frac{68}{31185} r - \frac{4}{2835} r^2 \right) x^{10}, \\ \frac{1}{3} + \frac{2}{3} \left( \frac{\sinh x}{x} \right)^{\frac{9r}{2}} (\cosh x)^{-\frac{3r}{2}} > 1 + \frac{1}{15} r x^4 + \frac{4}{189} r x^6 + \left( \frac{1}{150} r + \frac{1}{300} r^2 \right) x^8 + \\ \left( \frac{68}{31185} r + \frac{2}{945} r^2 \right) x^{10} + \left( \frac{42842}{58046625} r + \frac{1}{9000} r^3 + \frac{5969}{5953500} r^2 \right) x^{12}. $$

After regrouping and simplifying and using *Maple* it yields

$$ \frac{1}{3} + \frac{2}{3} \left( \frac{\sinh x}{x} \right)^{\frac{9r}{2}} (\cosh x)^{-\frac{3r}{2}} - \left( \frac{1}{\cosh x} \left( \frac{\sinh x}{x} \right)^3 \right)^r > $$
$$ \left( \frac{42842}{58046625} r + \frac{1}{9000} r^3 + \frac{5969}{5953500} r^2 \right) x^{12} - \frac{2}{2835} x^{10} r^2 + \frac{1}{900} x^8 r^2 > 0. $$

(iii) - Turn now to the last part of Theorem 1.5

A series representation of the left side with respect to $r$ yields

$$ \left( 3 - 2 \left( \frac{\sinh x}{x} \right)^{-3/2} (\cosh x)^{1/2} r \right)^{r^{-1}} = $$

$$ \frac{(\sinh x)^3}{x^3 \cos x} \left( 1 - \frac{3}{4} \left( 3 \ln \left( \frac{\sinh x}{x} \right) + \ln (\cosh x) \right)^2 r \right) + O(r^2). $$

This means

$$ \lim_{r \to 0} \left( 3 - 2 \left( \frac{\sinh x}{x} \right)^{-3/2} (\cosh x)^{1/2} r \right)^{r^{-1}} = \left( \frac{\sinh x}{x} \right)^3 \frac{1}{\cosh x}. $$

For the right side one gets the limited expansion with respect to $r$

$$ \left( 1/3 + 2/3 \left( \frac{\sinh x}{x} \right)^{9/2} (\cosh x)^{-3/2} \right)^{r^{-1}} = $$

$$ \frac{(\sinh x)^3}{x^3 \cosh x} \left( 1 + \frac{1}{4} \left( 3 \ln \left( \frac{\sinh x}{x} \right) + \ln (\cosh x) \right)^2 r \right) + O(r^2). $$

This means

$$ \lim_{r \to 0} \left( 1/3 + 2/3 \left( \frac{\sinh x}{x} \right)^{9/2} (\cosh x)^{-3/2} \right)^{r^{-1}} = \frac{(\sinh x)^3}{x^3 \cosh x}. $$

## 4 Links with older inequalities

In this section, we will more closely verify why the bounds given by Theorem 1.2 are the most precise concerning the Sinc hyperbolic function. As mentioned in the introduction, we will determine the value of $r_0$ for certain cases in order that for $r < r_0$ one has

$$ f(x) < \left( 3 - 2 \left( \frac{\sinh x}{x} \right)^{-\frac{r}{2}} \right)^{\frac{1}{r}}, \quad g(x) > \left( \frac{2}{3} \left( \frac{\sinh x}{x} \right)^{\frac{3r}{2}} + \frac{1}{3} \right)^{\frac{1}{r}}. $$

To do this, we will return to the examples cited in the introduction and let us compute in each case the difference thanks to Maple.
## 4.1 Inequality of Chen-Sandor [6]

It is easy to ensure that for $x \ge 0$ the following hold

$$ \left(3 - 2\left(\frac{\sinh x}{x}\right)^{\frac{-r}{2}}\right)^{\frac{1}{r}} - \left(\frac{1 + 2\cosh x}{3}\right)^{1/2} > 0 $$

as soon as $r < \frac{1}{10}$. This means

$$ \left(\frac{1 + 2 \cosh x}{3}\right)^{1/2} < \left(3 - 2\left(\frac{\sinh x}{x}\right)^{\frac{-r}{2}}\right)^{\frac{1}{r}} < \frac{\sinh x}{x} $$

## 4.2 Inequality of Bhayo-Sandor: Theorem 3.4 of [1]

We may easily verify that for $x \ge 0$ the following hold

$$ \left(3 - 2\left(\frac{\sinh x}{x}\right)^{\frac{-r}{2}}\right)^{\frac{1}{r}} - \left(\frac{1}{2} + \frac{1}{2}\cosh x\right)^{\frac{2}{3}} > 0 $$

as soon as $r < \frac{1}{70}$, which means

$$ \left(\frac{1}{2} + \frac{1}{2}\cosh x\right)^{\frac{2}{3}} < \left(3 - 2\left(\frac{\sinh x}{x}\right)^{\frac{-r}{2}}\right)^{\frac{1}{r}} < \frac{\sinh x}{x} $$

## 4.3 Left inequality of Bhayo-Sandor: Theorem 1.5 of [1]

It is easy to see that for $x \ge 0$ the following hold

$$ \left(3 - 2\left(\frac{\sinh x}{x}\right)^{\frac{-r}{2}}\right)^{\frac{1}{r}} - \exp\left(\frac{1}{2}\left(\frac{x}{\tanh x} - 1\right)\right) > 0 $$

as soon as $r < \frac{1}{70}$, that is to say

$$ \exp\left(\frac{1}{2}\left(\frac{x}{\tanh x} - 1\right)\right) < \left(3 - 2\left(\frac{\sinh x}{x}\right)^{\frac{-r}{2}}\right)^{\frac{1}{r}} < \frac{\sinh x}{x} $$

## 4.4 Right inequality of Bhayo-Sandor: Theorem 1.5 of [1]

We easily verify for $x \ge 0$ the following hold

$$ \left(\frac{2}{3}\left(\frac{\sinh x}{x}\right)^{\frac{3r}{2}} + \frac{1}{3}\right)^{\frac{1}{r}} - \exp\left(\left(\frac{x}{\tanh x} - 1\right)\right) < 0 $$

as soon as $r < \frac{1}{150}$, in other words

$$ \frac{\sinh x}{x} < \left(\frac{2}{3}\left(\frac{\sinh x}{x}\right)^{\frac{3r}{2}} + \frac{1}{3}\right)^{\frac{1}{r}} < \exp\left(\left(\frac{x}{\tanh x} - 1\right)\right). $$
## 4.5 Inequality of Wu-Li-Bencze [12]

Thanks to Maple it is easy to ensure that for $x \ge 0$ the following hold

$$ \left(3 - 2\left(\frac{\sinh x}{x}\right)^{\frac{-r}{2}}\right)^{\frac{1}{r}} - \cosh x - \frac{12}{5}\left(1 - \frac{x}{\sinh x}\right)^2 > 0 $$

as soon as $r < \frac{1}{100}$. This means

$$ \cosh x + \frac{12}{5}\left(1 - \frac{x}{\sinh x}\right)^2 < \left(3 - 2\left(\frac{\sinh x}{x}\right)^{\frac{-r}{2}}\right)^{\frac{1}{r}} < \frac{\sinh x}{x} $$

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